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- First point
- test
- Second one
- how to separate these two

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- Todos:
- LATER [#C] 买一台笔记本电脑
SCHEDULED: <2023-06-18 Sun>
:LOGBOOK:
CLOCK: [2023-03-26 Sun 15:39:27]--[2023-03-26 Sun 15:39:28] => 00:00:01
CLOCK: [2023-03-26 Sun 15:39:28]--[2023-03-26 Sun 15:39:29] => 00:00:01
CLOCK: [2023-03-26 Sun 15:39:30]--[2023-03-26 Sun 15:39:30] => 00:00:00
CLOCK: [2023-03-26 Sun 16:30:28]--[2023-03-26 Sun 16:30:28] => 00:00:00
:END:
- 配置：
- 酷睿 13／amdR9
- RTX4060
- 32+1T， 可替换内存和硬盘
- 16 英寸及以下
- 2kg 及以下
- 价格：
- 10000 左右
- 备选项
- | 品牌 | 尺寸 | 重量 | CPU | GPU | 内存 | 硬盘 | 内存拓展 | 硬盘拓展 | 价格 |
| ---- | ---- | ---- | ----- | ---- | ---- | ---- | -------- | -------- | ---------------------------------------- |
| 雷蛇 | 15 | 2kg | i7-13 | 4060 | 32 | 1T | 64 | 4T | ((64224ad1-0983-437e-89bc-779520a862aa)) |
- DONE 整理笔记
:LOGBOOK:
CLOCK: [2023-03-26 Sun 15:39:51]--[2023-03-26 Sun 15:39:54] => 00:00:03
CLOCK: [2023-03-26 Sun 15:39:55]--[2023-03-26 Sun 15:48:47] => 00:08:52
:END:
- LATER 修网站 #projects
- 前端
- 服务器
- DONE 学概率论
-
- Platform - Name
template:: OJ Problem Notes
- Times:
- Time when completed: <% time %>
- Time taken to complete:
- Revisions:
- Tags:
- Algorithms: to-be-filled
- Data structures: to-be-filled
- Difficulty: to-be-filled
- Platforms: to-be-filled
- Links:
- link to the problem
- Problem:
- to-be-filled
- Examples:
- ```
```
- Constraints:
- to-be-filled
- Thoughts:
- Intuition:
- Approach:
- Solution:
- Code
-
- [[draws/2023-03-26-16-47-17.excalidraw]]
Testing whiteboard
-
-

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- Todos:
- 学习任务
- DONE 交区块链作业
- DONE [#A] 剪视频
SCHEDULED: <2023-03-29 Wed>
DEADLINE: <2023-04-06 Thu>
:LOGBOOK:
CLOCK: [2023-03-28 Tue 09:54:49]--[2023-03-28 Tue 09:54:53] => 00:00:04
CLOCK: [2023-03-28 Tue 10:42:56]--[2023-03-28 Tue 11:57:08] => 01:14:12
CLOCK: [2023-03-28 Tue 12:48:27]--[2023-03-28 Tue 13:45:10] => 00:56:43
CLOCK: [2023-03-31 Fri 22:24:12]--[2023-03-31 Fri 23:42:11] => 01:17:59
CLOCK: [2023-04-01 Sat 10:42:09]--[2023-04-01 Sat 11:55:33] => 01:13:24
:END:
- DONE 学 java
:LOGBOOK:
CLOCK: [2023-03-27 Mon 11:26:38]--[2023-03-27 Mon 14:04:31] => 02:37:53
CLOCK: [2023-03-27 Mon 15:54:27]--[2023-03-27 Mon 16:02:50] => 00:08:23
CLOCK: [2023-03-27 Mon 16:02:52]--[2023-03-27 Mon 16:02:52] => 00:00:00
:END:
- DONE 概率论作业
SCHEDULED: <2023-04-01 Sat>
DEADLINE: <2023-04-01 Sat>
:LOGBOOK:
CLOCK: [2023-04-01 Sat 19:33:00]--[2023-04-01 Sat 22:53:21] => 03:20:21
:END:
- DONE 17
:LOGBOOK:
CLOCK: [2023-04-01 Sat 19:33:05]--[2023-04-01 Sat 20:48:19] => 01:15:14
:END:
- DONE 18 密度改成 1 / 4 pi
- DONE 20
:LOGBOOK:
CLOCK: [2023-04-01 Sat 20:48:22]--[2023-04-01 Sat 20:55:58] => 00:07:36
:END:
- DONE 22
:LOGBOOK:
CLOCK: [2023-04-01 Sat 20:55:59]--[2023-04-01 Sat 21:08:03] => 00:12:04
:END:
- DONE 54
:LOGBOOK:
CLOCK: [2023-04-01 Sat 21:08:04]--[2023-04-01 Sat 22:53:17] => 01:45:13
:END:
- DONE 55
- DONE 58 b 不做
- DONE 59
- DONE 复习 java
SCHEDULED: <2023-03-28 Tue>
- DONE 做 lab
:LOGBOOK:
CLOCK: [2023-03-27 Mon 18:59:33]--[2023-03-27 Mon 21:51:19] => 02:51:46
:END:
- DONE 产开选题
DEADLINE: <2023-03-30 Thu>
- DONE [#A] 复习电路
DEADLINE: <2023-03-29 Wed>
:LOGBOOK:
CLOCK: [2023-03-28 Tue 09:52:06]--[2023-03-28 Tue 10:59:57] => 01:07:51
:END:
- DONE 看课件
:LOGBOOK:
CLOCK: [2023-03-28 Tue 09:54:34]--[2023-03-28 Tue 10:41:52] => 00:47:18
:END:
- DONE 做题
- DONE [#A] 复习卡氏图
:LOGBOOK:
CLOCK: [2023-03-29 Wed 10:31:01]--[2023-03-29 Wed 10:31:01] => 00:00:00
CLOCK: [2023-03-29 Wed 10:31:02]--[2023-03-29 Wed 16:52:13] => 06:21:11
:END:
- DONE 预习 lab
- DONE 打印资料
- 个人任务
- DONE 给尧舜天注册帐号 (没注册成功)
:LOGBOOK:
CLOCK: [2023-03-27 Mon 16:13:04]--[2023-03-27 Mon 16:41:33] => 00:28:29
CLOCK: [2023-03-27 Mon 17:58:01]--[2023-03-27 Mon 17:58:01] => 00:00:00
:END:
- LATER Work on mapping feature in logseq
- DONE 洗衣服
SCHEDULED: <2023-03-28 Tue>
:LOGBOOK:
CLOCK: [2023-03-28 Tue 08:53:36]--[2023-03-28 Tue 09:37:35] => 00:43:59
:END:
-
- Recurring tasks: [[Habits]]
- Daily:
- LATER [#B] Read books
SCHEDULED: <2023-05-24 Wed .+1d>
:LOGBOOK:
CLOCK: [2023-03-27 Mon 16:47:35]--[2023-03-27 Mon 16:47:51] => 00:00:16
- State "DONE" from "NOW" [2023-03-27 Mon 16:47]
CLOCK: [2023-03-30 Thu 08:04:21]--[2023-03-30 Thu 09:16:49] => 01:12:28
* State "DONE" from "LATER" [2023-03-30 Thu 09:16]
CLOCK: [2023-04-02 Sun 19:59:18]--[2023-04-02 Sun 20:17:05] => 00:17:47
* State "DONE" from "LATER" [2023-04-04 Tue 14:33]
* State "DONE" from "LATER" [2023-04-06 Thu 07:33]
* State "DONE" from "LATER" [2023-04-10 Mon 18:33]
* State "DONE" from "LATER" [2023-04-12 Wed 16:17]
* State "DONE" from "LATER" [2023-04-15 Sat 16:00]
* State "DONE" from "LATER" [2023-04-15 Sat 20:37]
* State "DONE" from "LATER" [2023-04-18 Tue 09:55]
* State "DONE" from "LATER" [2023-04-19 Wed 18:10]
* State "DONE" from "LATER" [2023-04-21 Fri 15:14]
CLOCK: [2023-04-22 Sat 19:10:58]--[2023-04-22 Sat 21:39:19] => 02:28:21
* State "DONE" from "NOW" [2023-04-22 Sat 21:39]
CLOCK: [2023-04-23 Sun 09:37:33]--[2023-04-23 Sun 10:59:35] => 01:22:02
* State "DONE" from "NOW" [2023-04-23 Sun 10:59]
* State "DONE" from "LATER" [2023-04-24 Mon 21:39]
* State "DONE" from "LATER" [2023-04-26 Wed 22:31]
* State "DONE" from "LATER" [2023-04-28 Fri 17:31]
CLOCK: [2023-05-06 Sat 20:35:36]--[2023-05-06 Sat 21:40:06] => 01:04:30
* State "DONE" from "NOW" [2023-05-06 Sat 21:40]
* State "DONE" from "LATER" [2023-05-08 Mon 20:15]
* State "DONE" from "LATER" [2023-05-09 Tue 11:13]
* State "DONE" from "LATER" [2023-05-10 Wed 21:35]
* State "DONE" from "LATER" [2023-05-11 Thu 19:55]
* State "DONE" from "LATER" [2023-05-15 Mon 15:43]
* State "DONE" from "LATER" [2023-05-20 Sat 20:10]
* State "DONE" from "LATER" [2023-05-23 Tue 20:26]
:END:
- LATER [#B] Memorize vocabulary
SCHEDULED: <2023-05-24 Wed .+1d>
:LOGBOOK:
CLOCK: [2023-03-27 Mon 16:48:26]--[2023-03-27 Mon 18:53:54] => 02:05:28
- State "DONE" from "NOW" [2023-03-27 Mon 18:53]
* State "DONE" from "LATER" [2023-04-02 Sun 14:39]
* State "DONE" from "LATER" [2023-04-03 Mon 10:30]
* State "DONE" from "LATER" [2023-04-06 Thu 22:51]
* State "DONE" from "LATER" [2023-04-11 Tue 13:52]
* State "DONE" from "LATER" [2023-04-12 Wed 12:29]
* State "DONE" from "LATER" [2023-04-13 Thu 21:42]
* State "DONE" from "LATER" [2023-04-15 Sat 17:34]
CLOCK: [2023-04-17 Mon 09:43:03]--[2023-04-17 Mon 15:57:16] => 06:14:13
* State "DONE" from "NOW" [2023-04-17 Mon 15:57]
* State "DONE" from "LATER" [2023-04-18 Tue 10:13]
CLOCK: [2023-04-23 Sun 11:10:16]--[2023-04-23 Sun 11:28:28] => 00:18:12
* State "DONE" from "NOW" [2023-04-23 Sun 11:28]
* State "DONE" from "LATER" [2023-04-24 Mon 20:57]
* State "DONE" from "LATER" [2023-04-25 Tue 10:57]
* State "DONE" from "LATER" [2023-04-26 Wed 22:31]
* State "DONE" from "LATER" [2023-04-28 Fri 16:37]
* State "DONE" from "LATER" [2023-05-06 Sat 20:35]
* State "DONE" from "LATER" [2023-05-07 Sun 09:20]
* State "DONE" from "LATER" [2023-05-08 Mon 19:36]
CLOCK: [2023-05-09 Tue 09:50:18]--[2023-05-09 Tue 09:52:45] => 00:02:27
* State "DONE" from "NOW" [2023-05-09 Tue 09:52]
CLOCK: [2023-05-11 Thu 10:13:10]--[2023-05-11 Thu 10:13:11] => 00:00:01
* State "DONE" from "NOW" [2023-05-11 Thu 10:13]
* State "DONE" from "LATER" [2023-05-15 Mon 15:43]
* State "DONE" from "LATER" [2023-05-19 Fri 16:55]
* State "DONE" from "LATER" [2023-05-21 Sun 20:59]
* State "DONE" from "LATER" [2023-05-23 Tue 20:26]
:END:
- DONE [#B] Solve one #leetcode Problem
SCHEDULED: <2023-05-22 Mon>
:LOGBOOK:
CLOCK: [2023-03-27 Mon 16:56:53]--[2023-03-28 Tue 14:02:38] => 21:05:45
CLOCK: [2023-03-28 Tue 14:02:43]--[2023-03-28 Tue 14:24:54] => 00:22:11
- State "DONE" from "NOW" [2023-03-28 Tue 14:24]
CLOCK: [2023-03-30 Thu 07:39:35]--[2023-03-30 Thu 07:52:53] => 00:13:18
* State "DONE" from "NOW" [2023-03-30 Thu 07:52]
CLOCK: [2023-03-31 Fri 07:51:59]--[2023-03-31 Fri 08:41:05] => 00:49:06
* State "DONE" from "NOW" [2023-03-31 Fri 08:41]
CLOCK: [2023-04-02 Sun 09:36:38]--[2023-04-02 Sun 10:35:35] => 00:58:57
* State "DONE" from "LATER" [2023-04-02 Sun 10:35]
CLOCK: [2023-04-03 Mon 18:56:00]--[2023-04-03 Mon 22:00:42] => 03:04:42
* State "DONE" from "NOW" [2023-04-03 Mon 22:00]
CLOCK: [2023-04-06 Thu 07:33:36]--[2023-04-06 Thu 07:36:41] => 00:03:05
* State "DONE" from "LATER" [2023-04-06 Thu 10:21]
* State "DONE" from "LATER" [2023-04-11 Tue 10:35]
CLOCK: [2023-04-16 Sun 10:11:02]--[2023-04-17 Mon 09:43:05] => 23:32:03
* State "DONE" from "LATER" [2023-04-17 Mon 22:58]
* State "DONE" from "LATER" [2023-04-18 Tue 09:55]
CLOCK: [2023-04-21 Fri 14:11:20]--[2023-04-21 Fri 14:41:02] => 00:29:42
* State "DONE" from "NOW" [2023-04-21 Fri 14:41]
* State "DONE" from "LATER" [2023-04-25 Tue 10:46]
* State "DONE" from "LATER" [2023-04-28 Fri 16:45]
* State "DONE" from "LATER" [2023-05-11 Thu 20:30]
* State "DONE" from "LATER" [2023-05-21 Sun 20:59]
:END:
- Weekly:
- LATER [#B] Run in the field three times a week 2/3
:LOGBOOK:
CLOCK: [2023-03-27 Mon 16:58:41]--[2023-03-27 Mon 19:33:31] => 02:34:50
- State "DONE" from "NOW" [2023-03-27 Mon 19:33]
* State "DONE" from "LATER" [2023-03-28 Tue 19:35]
* State "DONE" from "LATER" [2023-03-30 Thu 07:30]
* State "DONE" from "LATER" [2023-03-31 Fri 07:52]
* State "DONE" from "LATER" [2023-04-02 Sun 09:35]
* State "DONE" from "LATER" [2023-04-07 Fri 11:32]
* State "DONE" from "LATER" [2023-04-09 Sun 16:57]
* State "DONE" from "LATER" [2023-04-11 Tue 12:47]
* State "DONE" from "LATER" [2023-04-12 Wed 08:59]
* State "DONE" from "LATER" [2023-04-13 Thu 21:42]
* State "DONE" from "LATER" [2023-04-14 Fri 08:37]
* State "DONE" from "LATER" [2023-04-15 Sat 16:00]
* State "DONE" from "LATER" [2023-04-18 Tue 23:31]
* State "DONE" from "LATER" [2023-04-19 Wed 22:05]
* State "DONE" from "LATER" [2023-04-21 Fri 09:07]
* State "DONE" from "LATER" [2023-04-23 Sun 13:31]
* State "DONE" from "LATER" [2023-04-24 Mon 20:56]
* State "DONE" from "LATER" [2023-04-25 Tue 11:24]
* State "DONE" from "LATER" [2023-04-28 Fri 16:18]
* State "DONE" from "LATER" [2023-05-06 Sat 20:25]
* State "DONE" from "LATER" [2023-05-08 Mon 19:25]
* State "DONE" from "LATER" [2023-05-10 Wed 21:35]
* State "DONE" from "LATER" [2023-05-11 Thu 19:55]
* State "DONE" from "LATER" [2023-05-12 Fri 08:38]
CLOCK: [2023-05-19 Fri 17:49:26]--[2023-05-19 Fri 17:49:28] => 00:00:02
* State "DONE" from "LATER" [2023-05-19 Fri 17:49]
* State "DONE" from "LATER" [2023-05-21 Sun 20:59]
:END:
-
- LATER [#C] Watch movies once a week
SCHEDULED: <2023-06-12 Mon .+2d>
:LOGBOOK:
CLOCK: [2023-03-27 Mon 16:59:18]--[2023-03-27 Mon 19:34:08] => 02:34:50
- State "DONE" from "NOW" [2023-03-27 Mon 19:34]
* State "DONE" from "LATER" [2023-04-10 Mon 22:13]
* State "DONE" from "LATER" [2023-04-12 Wed 16:00]
* State "DONE" from "LATER" [2023-04-14 Fri 08:37]
* State "DONE" from "LATER" [2023-04-18 Tue 23:31]
* State "DONE" from "LATER" [2023-04-21 Fri 09:07]
* State "DONE" from "LATER" [2023-04-22 Sat 19:10]
* State "DONE" from "LATER" [2023-04-24 Mon 20:56]
* State "DONE" from "LATER" [2023-05-06 Sat 20:25]
* State "DONE" from "LATER" [2023-05-08 Mon 19:25]
* State "DONE" from "LATER" [2023-05-11 Thu 20:08]
* State "DONE" from "LATER" [2023-05-14 Sun 13:15]
* State "DONE" from "LATER" [2023-05-19 Fri 17:35]
* State "DONE" from "LATER" [2023-06-08 Thu 11:29]
* State "DONE" from "LATER" [2023-06-10 Sat 16:37]
:END:
-
-
- 每周学习任务
- 每周:
- 智慧树网课
:LOGBOOK:
CLOCK: [2023-03-27 Mon 17:22:00]
:END:
- DONE [#C] 网课每周三次 3/3
:LOGBOOK:
CLOCK: [2023-03-27 Mon 17:23:09]--[2023-03-27 Mon 17:23:18] => 00:00:09
- State "DONE" from "NOW" [2023-03-27 Mon 17:23]
- State "DONE" from "LATER" [2023-03-27 Mon 17:24]
- State "DONE" from "LATER" [2023-03-27 Mon 17:25]
- State "DONE" from "LATER" [2023-03-27 Mon 19:27]
CLOCK: [2023-03-27 Mon 19:29:22]--[2023-03-27 Mon 19:29:31] => 00:00:09
- State "DONE" from "NOW" [2023-03-27 Mon 19:29]
* State "DONE" from "LATER" [2023-03-28 Tue 19:35]
* State "DONE" from "LATER" [2023-03-29 Wed 16:52]
* State "DONE" from "LATER" [2023-03-30 Thu 09:16]
* State "DONE" from "LATER" [2023-03-31 Fri 07:52]
* State "DONE" from "LATER" [2023-04-02 Sun 09:35]
* State "DONE" from "LATER" [2023-04-03 Mon 13:31]
CLOCK: [2023-04-03 Mon 22:02:50]--[2023-04-04 Tue 09:09:23] => 11:06:33
:END:
- DONE [#C] 测验每周一次 1/1
:LOGBOOK:
CLOCK: [2023-03-27 Mon 17:23:12]--[2023-03-27 Mon 19:27:27] => 02:04:15
- State "DONE" from "NOW" [2023-03-27 Mon 19:27]
CLOCK: [2023-03-27 Mon 19:29:26]--[2023-03-27 Mon 19:29:35] => 00:00:09
- State "DONE" from "NOW" [2023-03-27 Mon 19:29]
* State "DONE" from "LATER" [2023-03-29 Wed 18:28]
* State "DONE" from "LATER" [2023-03-30 Thu 09:16]
* State "DONE" from "LATER" [2023-04-01 Sat 09:25]
* State "DONE" from "LATER" [2023-04-02 Sun 09:35]
:END:
- DONE [#C] 互动每周三次 1/3
:LOGBOOK:
CLOCK: [2023-03-27 Mon 17:23:16]--[2023-03-27 Mon 19:29:36] => 02:06:20
- State "DONE" from "NOW" [2023-03-27 Mon 19:29]
* State "DONE" from "LATER" [2023-03-31 Fri 11:16]
CLOCK: [2023-04-02 Sun 19:10:30]--[2023-04-02 Sun 19:25:28] => 00:14:58
* State "DONE" from "NOW" [2023-04-02 Sun 19:25]
* State "DONE" from "LATER" [2023-04-03 Mon 18:49]
CLOCK: [2023-04-04 Tue 13:35:09]--[2023-04-04 Tue 13:55:20] => 00:20:11
* State "DONE" from "NOW" [2023-04-04 Tue 13:55]
* State "DONE" from "LATER" [2023-04-06 Thu 22:51]
* State "DONE" from "LATER" [2023-04-11 Tue 16:27]
:END:
-
- [#C] 这几天的书单 [[Book lists]] #excluded_in_query #books
- 通识
- DONE [#C] 云彩收集者手册
:LOGBOOK:
CLOCK: [2023-03-27 Mon 19:41:38]
CLOCK: [2023-03-27 Mon 19:42:00]--[2023-03-28 Tue 16:19:55] => 20:37:55
CLOCK: [2023-03-28 Tue 16:20:29]--[2023-03-31 Fri 07:46:16] => 63:25:47
:END:
- NOW [#C] 日本料理完全图鉴
:LOGBOOK:
CLOCK: [2023-03-27 Mon 19:42:14]
:END:
- 历史
- DONE [#C] 二手时间
:LOGBOOK:
CLOCK: [2023-03-27 Mon 19:42:23]--[2023-04-15 Sat 18:53:54] => 455:11:31
CLOCK: [2023-04-15 Sat 18:53:55]--[2023-04-15 Sat 18:53:55] => 00:00:00
:END:
- 文学
- LATER the eternaut
- 经典小说
- LATER 欧亨利
- LATER 莫泊桑
- LATER 契诃夫
- LATER 奥勃洛莫夫
- 爱情小说
- LATER Price and prejudice
- 哲学
- LATER How to be a stoic
- DONE [[Amused to Death]]
:LOGBOOK:
CLOCK: [2023-03-31 Fri 07:47:22]
CLOCK: [2023-03-31 Fri 07:47:34]--[2023-04-18 Tue 09:54:03] => 434:06:29
:END:
- DONE [[亲密关系]]
:LOGBOOK:
CLOCK: [2023-04-15 Sat 18:54:13]--[2023-04-23 Sun 10:52:43] => 183:58:30
:END:
- DONE [[拥抱逝水年华]]
-
- Book note [[Book notes]]
template:: Book Note
collapsed:: true
- Metadata:
- Time: <% time %>
- Page: optional
- Content:
- This is pretty good
-
- Daily reflection [[Daily reflections]]
template:: Daily reflections
collapsed:: true
- What I've done
- What I've thought #thoughts
- Mood
-
- Cloud collections [[Cloud collections]]
template:: Cloud Collections
collapsed:: true
- Photo
- Notes
- Metadata
- Time
- <% time %> <% today %>
- Weather
- cloudy
- Type
- Location
- Map
<iframe src="https://www.google.com/maps/embed?pb=!1m18!1m12!1m3!1d3049.2920871146916!2d116.28518!3d40.158049999999996!2m3!1f0!2f0!3f0!3m2!1i1024!2i768!4f13.1!3m3!1m2!1s0x35f0fbce3d955439%3A0x767d9f097c49f8eb!2sBupt%20Shahe%20Campus!5e0!3m2!1sen!2sjp!4v1679910773010!5m2!1sen!2sjp" width="600" height="450" style="border:0;" allowfullscreen="" loading="lazy" referrerpolicy="no-referrer-when-downgrade"></iframe>
- Coordination
- 40.15805,116.28518
-
- #joke
- Gob 林

245
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@ -0,0 +1,245 @@
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- What I've thought #thoughts
- 10:08
- logseq 比 obsidian 好用，计笔记成本低，而且好用，便携。
- Mood
- 10:09
- 平静且开心(lol)
-
- DONE [#C] 票在书包里
SCHEDULED: <2023-03-30 Thu>
-
- Todos:
- Personal
- DONE 重新浸洗袜子和内裤
SCHEDULED: <2023-03-28 Tue 13:00>
:LOGBOOK:
CLOCK: [2023-03-28 Tue 15:55:30]--[2023-03-28 Tue 15:55:31] => 00:00:01
:END:
- DONE 洗眼镜
SCHEDULED: <2023-03-28 Tue 13:00>
:LOGBOOK:
CLOCK: [2023-03-28 Tue 18:07:51]--[2023-03-28 Tue 22:41:19] => 04:33:28
:END:
- DONE Anki card for cloud collection 还不如没事多翻下书
SCHEDULED: <2023-04-01 Sat>
- School
- DONE 交选题
:LOGBOOK:
CLOCK: [2023-03-28 Tue 16:19:35]
CLOCK: [2023-03-28 Tue 16:19:40]--[2023-03-28 Tue 18:51:31] => 02:31:51
:END:
- DONE [#A] 复习 java
:LOGBOOK:
CLOCK: [2023-03-28 Tue 20:03:20]--[2023-03-29 Wed 22:25:48] => 26:22:28
:END:
- DONE 做题
:LOGBOOK:
CLOCK: [2023-03-28 Tue 20:03:29]--[2023-03-29 Wed 11:11:16] => 15:07:47
:END:
- DONE 练习题
:LOGBOOK:
CLOCK: [2023-03-28 Tue 20:03:51]--[2023-03-28 Tue 20:39:32] => 00:35:41
:END:
- DONE 往年题
:LOGBOOK:
CLOCK: [2023-03-28 Tue 20:03:50]--[2023-03-28 Tue 22:41:14] => 02:37:24
:END:
- DONE Leetcode
:LOGBOOK:
CLOCK: [2023-03-29 Wed 11:19:02]
CLOCK: [2023-03-29 Wed 11:19:04]--[2023-03-29 Wed 16:52:09] => 05:33:05
:END:
- DONE 看 ppt
:LOGBOOK:
CLOCK: [2023-03-28 Tue 20:03:30]--[2023-03-29 Wed 11:18:53] => 15:15:23
:END:
-
- OJ problem notes
- Name:
- Contains Duplicate
- Times:
- Completion time: 14:31
- Revisions:
- Tags:
- Data structures: #hash_table
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/contains-duplicate/description/)
- Problem:
- Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.
- Examples:
- ```
Example 1:
Input: nums = [1,2,3,1]
Output: true
Example 2:
Input: nums = [1,2,3,4]
Output: false
Example 3:
Input: nums = [1,1,1,3,3,4,3,2,4,2]
Output: true
```
- Constraints:
- `1 <= nums.length <= 105`
- `-109 <= nums[i] <= 109`
- Thoughts:
- Very simple hash map problem
- Solution:
- ```java
class Solution {
public boolean containsDuplicate(int[] nums) {
Set <Integer> set = new HashSet<Integer>();
for (int i : nums) {
if (set.contains(i)) {
return true;
} else {
set.add(i);
}
}
return false;
}
}
```
-
-
-
- Java 练习题 1 #to_be_deleted
collapsed:: true
- Q1
- `less than 10`
- Q2
- `greater than 10`
- Q3
- `yes`
- Q4
- ```
zero
one
none
```
- Q5
- ```
zero
one
none
one
none
none
```
- Q6
- ```
zero
none
one
zero
none
none
```
- Q7
- ```java
int sum;
for (int i = 1; i <= 100; i++) {
sum += i;
}
System.out.println(sum);
```
- Q8
- ```java
for (int i = 1; i <= 3; i++) {
for (int j = 1; j <= 3; j++) {
System.out.print(i * j + " ");
}
System.out.println();
}
```
- Q9
- ```
9
8
true
```
- Q10
- ```java
int sum = 0;
for (int i = 1; i <= 99; i++) {
if (i % 3 == 0) {
sum += i;
}
}
System.out.println(sum);
```
- Q11 不想写
- Q12
- What does overloading mean?
- What does overloading do?
-
- java 练习题 2 [qmplus](https://qmplus.qmul.ac.uk/course/view.php?id=21582#section-8) #to_be_deleted
- Q1
- no.1 : not a array
no.2 : `int[] marks = new int[60]`
no.3 : `{'a', 'b', 'c'}`
no.4 : use ""
- Q2
- Compile error
- Q3
- `F[1]` is not initiated, it only has a null value
- Q4
- Array index out of bounds, f[2] is non-exist
- Q5
- compile error, local variable won't be initialized as default val
- Q6
- 36, 36.0
- Q7
- first one invalid
- Reason: wrong parameters
- third one invalid
- Reason: double->int can't be done implicitly, you have to use type casts
- Q8
- ```
vroom
truck 1
car 2
```
- Q9
- ```
vroomT
truck 1
car 1
```
- Q10
- 634.5
- Reason: calculated from left to right, all converted to String
- Q11
- F, only have access to package-private data
- F, no multi-inheritance for classes
- F, not required
- F, you can have one
- Q12
- b
- Q13
- false
- true
- true (reference stored in string constant pool)
- Q14
collapsed:: true
- 懒得写了
- Q15
collapsed:: true
- 懒得写
- Q16
- What is a abstract class?
- What is `toString` method?
- What does polymorphism mean?

264
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- #projects [[New computer 2023]] Gnome vs. Window managers
- Gnome
- pro
- looks beautiful by default
- good touch pad gestures
- con
- keyboard centered workflow requires additional config.
- [paperWM](https://github.com/paperwm/PaperWM)
- [Pop-os shell](https://github.com/pop-os/shell) might not support gnome 43
- forge
- Window managers like sway
- pro
- keyboard-centered workflow is fast and suits laptops well
- looks good [unixporn](https://www.reddit.com/r/unixporn)
- fun to use and configure
- con
- have to configure many stuff by hand
collapsed:: true
- 1. Multi-screen setup
2. volume management that looks decent
3. various xdg-desktop portals
- Color management
- automatic theme switching and night mode configuration
- sway and Hyprland have bad support for nvidia
- Additional notes
- I think I will try sway and hypr TWM, and fall back to gnome. Use flatpak for large desktop applications
- [2022 status of nvidia's wayland driver on Sway and Gnome](https://blog.devgenius.io/wayland-and-nvidia-in-2022-2f0407fb34f4?gi=db52b330779c)
- When installing gnome, disable extra flag to reduce bloat
-
- [[New computer 2023]] Cool linux rices
- [SwayWM](https://www.reddit.com/r/unixporn/comments/124yjwy/minimalistic_sway_waybar_nixos_full_silent_boot/)
- [Hyprland](https://www.reddit.com/r/unixporn/comments/12315m9/hyprland_updating_dotfiles_again/) the bar looks pretty decent
-
- [[New computer 2023]] Stuff to remember
- Data in my home folder (Better back up them)
- SSH keys, TOTP keys
- Documents - Personal
- Flatpak steam stuff
- Gentoo specific
- gentoo-lto by default
- NVIDIA drivers
- Gnome/systemd profile (install gnome and gdm first then other WM)
-
- Todo
- Personal
- DONE Try putting my blogs on medium
SCHEDULED: <2023-04-01 Sat>
- DONE Theme flatpak apps
SCHEDULED: <2023-04-01 Sat>
-
-
- Daily reflection [[Daily reflections]]
- What I've done
- Review school stuff
- Didn't read books :(
- What I've thought #thoughts
- After watching youtube 1 hour straight, I feel some kind of bad.
My parents have been a hindrance to my personal development, and they
obviously don't want to commit that. They are bad, lame, unresponsive and in general disappointing.
- Mood
- For the most part great, because I feel spring is coming
-
- PittCS's Leetcode problem sets: #leetcode #excluded_in_query
id:: 6423af98-f01f-4c69-a2a3-6b38715bd077
collapsed:: true
- ### DONE Week 1 - Arrays
- DONE [Contains Duplicate](https://leetcode.com/problems/contains-duplicate/)
:LOGBOOK:
CLOCK: [2023-03-29 Wed 11:31:49]--[2023-03-29 Wed 11:35:13] => 00:03:24
:END:
- DONE [Two Sum](https://leetcode.com/problems/two-sum/) [[Leetcode Two-Sum]]
:LOGBOOK:
CLOCK: [2023-03-29 Wed 11:35:42]--[2023-03-29 Wed 11:49:32] => 00:13:50
:END:
- DONE [Buy & Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) [[Leetcode Best-Time-To-Buy-And-Sell-Stock]]
:LOGBOOK:
CLOCK: [2023-03-30 Thu 07:41:08]--[2023-03-30 Thu 07:51:53] => 00:10:45
:END:
- DONE [Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/) [[Leetcode Product-of-Array-Except-Self]]
:LOGBOOK:
CLOCK: [2023-03-30 Thu 07:53:22]--[2023-03-30 Thu 08:01:25] => 00:08:03
:END:
- DONE [Maximum Subarray](https://leetcode.com/problems/maximum-subarray/) [[Leetcode Maxinum-subarray]]
:LOGBOOK:
CLOCK: [2023-03-31 Fri 08:06:18]--[2023-03-31 Fri 08:15:11] => 00:08:53
:END:
- DONE [Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/) ((64261f45-6885-422c-808c-9344273af1d4))
:LOGBOOK:
CLOCK: [2023-03-31 Fri 08:23:01]--[2023-03-31 Fri 08:38:24] => 00:15:23
:END:
- DONE [Game of Life (**Bonus**)](https://leetcode.com/problems/game-of-life/) ((6428db9f-8f14-41a4-b5ec-5e79ffeb204a))
:LOGBOOK:
CLOCK: [2023-04-02 Sun 09:36:45]--[2023-04-02 Sun 09:36:50] => 00:00:05
CLOCK: [2023-04-02 Sun 09:50:33]--[2023-04-02 Sun 10:18:21] => 00:27:48
:END:
- #### DONE Week 1 - Arrays Resources
- DONE [Sets in Python](https://youtu.be/sBvaPopWOmQ)
- DONE [Dictionaries in Python](https://youtu.be/XCcpzWs-CI4)
- DONE [Traversing 2D Lists in Python](https://youtu.be/B5GhlXhDfoE)
- ### LATER Week 2 - Arrays + Intervals
:LOGBOOK:
CLOCK: [2023-04-02 Sun 14:57:17]--[2023-04-02 Sun 14:57:20] => 00:00:03
:END:
- DONE [Min Rotate Sort Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) ((64294296-85d5-4d21-b815-b4bad7525df5))
:LOGBOOK:
CLOCK: [2023-04-02 Sun 16:12:11]--[2023-04-02 Sun 16:53:34] => 00:41:23
:END:
- DONE [Search Rotate Sort Array](https://leetcode.com/problems/search-in-rotated-sorted-array/) ((642aa126-13d8-48ff-8eb4-f17d875add97))
:LOGBOOK:
CLOCK: [2023-04-02 Sun 17:14:27]--[2023-04-02 Sun 17:40:32] => 00:26:05
:END:
- DONE [3Sum](https://leetcode.com/problems/3sum/) ((642abd85-5c14-42c3-aa39-e7dc2773b2ec))
:LOGBOOK:
CLOCK: [2023-04-03 Mon 19:09:33]--[2023-04-03 Mon 19:50:58] => 00:41:25
:END:
- DONE [Container w Most Water](https://leetcode.com/problems/container-with-most-water/) ((642ac7bd-a908-4155-b839-f58ca64f4672))
:LOGBOOK:
CLOCK: [2023-04-03 Mon 19:55:22]--[2023-04-03 Mon 20:32:51] => 00:37:29
:END:
- DONE [Insert Interval](https://leetcode.com/problems/insert-interval/) ((642cd1a8-9707-4dc2-9d03-11464553e618))
:LOGBOOK:
CLOCK: [2023-04-05 Wed 10:37:25]--[2023-04-05 Wed 10:37:26] => 00:00:01
:END:
- DONE [Merge Intervals](https://leetcode.com/problems/merge-intervals/) ((642edfd2-c458-484e-9e94-8802771475d5))
:LOGBOOK:
CLOCK: [2023-04-06 Thu 08:56:14]--[2023-04-06 Thu 09:33:45] => 00:37:31
:END:
- DONE [Happy Number (**Bonus**)](https://leetcode.com/problems/happy-number/)
- #### DONE Week 2 - Arrays Resources
- LATER [How Binary Search Makes Computers Much, Much Faster](https://youtu.be/KXJSjte_OAI)
- LATER [Introduction to Binary Search](https://www.youtube.com/watch?v=6ysjqCUv3K4)
- LATER [Algorithms: Binary Search](https://www.youtube.com/watch?v=P3YID7liBug)
- ### LATER Week 3 - Intervals + Matrix
- DONE [Non-overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/) ((643d27e0-7330-4510-a2f7-8e2cecc08e7b))
- DONE [Meeting Rooms I](https://www.leetcode.com/problem/meeting-rooms/) This quesion is premium only
- DONE [Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)
- DONE [Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)
- DONE [Rotate Img](https://leetcode.com/problems/rotate-image/) ((64422d1f-5db3-435d-ab4e-2b9fe5fd47cd))
- LATER [Word Search](https://leetcode.com/problems/word-search/)
- LATER [Sliding Window Max (**Bonus**)](https://leetcode.com/problems/sliding-window-maximum/)
- #### LATER Week 3 - Matrix Resources
- LATER [2D Lists & Nested Loops - Python](https://youtu.be/B5GhlXhDfoE)
- ### LATER Week 4 - Linked Lists
- LATER [Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)
- LATER [Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)
- LATER [Merge 2 Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)
- LATER [Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)
- LATER [Remove Nth Node](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)
- LATER [Reorder List](https://leetcode.com/problems/reorder-list/)
- LATER [Sort Colors (**Bonus**)](https://leetcode.com/problems/sort-colors/)
- #### LATER Week 4 - Linked Lists Resources
- LATER [Data Structures: Linked Lists](https://youtu.be/njTh_OwMljA)
- LATER [Introduction to Linked Lists](https://youtu.be/WwfhLC16bis)
- LATER [Data Structures: Heaps](https://youtu.be/X4RWHH69Gi8)
- LATER [HeapQ for Algorithms in Python](https://youtu.be/X4RWHH69Gi8)
- ### LATER Week 5 - String
- LATER [Reverse String](https://leetcode.com/problems/reverse-string/)
- LATER [Valid Anagram](https://leetcode.com/problems/valid-anagram/)
- LATER [Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)
- LATER [Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)
- LATER [Group Anagrams](https://leetcode.com/problems/group-anagrams/)
- LATER [Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)
- LATER [Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)
- LATER [Odd Even Linked List (**Bonus**)](https://leetcode.com/problems/odd-even-linked-list/)
- #### LATER Week 5 - String Resources
- LATER [Python Stacks](https://youtu.be/NKmasqr_Xkw)
- ### LATER Week 6 - String + Trees
- LATER [Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)
- LATER [Longest Repeating Char Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/)
- LATER [Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)
- LATER [Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/)
- LATER [Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/)
- LATER [Tree Postorder Traversal](https://leetcode.com/problems/binary-tree-postorder-traversal/)
- LATER [Encode + Decode Strings](https://www.lintcode.com/problem/encode-and-decode-strings/)
- LATER [Max Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)
- LATER **Bonus** - Do all the Tree traversal problems iteratively
- LATER [Find Duplicate Number (**Bonus**)](https://leetcode.com/problems/find-the-duplicate-number/)
- #### LATER Week 6 - Trees Resources
- LATER [Data Structures: Trees](https://www.youtube.com/watch?v=oSWTXtMglKE)
- ### LATER Week 7 - Trees
- LATER [Same Tree](https://leetcode.com/problems/same-tree/)
- LATER [Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)
- LATER [Binary Tree Max Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)
- LATER [Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/)
- LATER [Serialize + Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)
- LATER [Subtree of Another Tre](https://leetcode.com/problems/subtree-of-another-tree/)
- LATER [LRU Cache (Bonus)](https://leetcode.com/problems/lru-cache/)
- ### LATER Week 8 - Trees
- LATER [Construct Binary Tree From Preorder and Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
- LATER [Validate BST](https://leetcode.com/problems/validate-binary-search-tree/)
- LATER [Kth Smallest Element in BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)
- LATER [LCA of BST](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)
- LATER [Implement Trie](https://leetcode.com/problems/implement-trie-prefix-tree/)
- LATER [Design Add + Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/)
- LATER [University Career Fair (**Bonus**)](https://cdn.discordapp.com/attachments/758353765174083595/840082064090595388/cHg7Bod.png)
- ### LATER Week 9 - Graphs
- LATER [Number of Islands](https://leetcode.com/problems/number-of-islands/)
- LATER [All Paths from Src](https://leetcode.com/problems/all-paths-from-source-to-target/)
- LATER [Word Search II](https://leetcode.com/problems/word-search-ii/)
- LATER [Clone Graph](https://leetcode.com/problems/clone-graph/)
- LATER [Course Schedule](https://leetcode.com/problems/course-schedule/)
- LATER [Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)
- LATER [Longest Consecutive Seq](https://leetcode.com/problems/longest-consecutive-sequence/)
- LATER [Longest Common Prefix (**Bonus**)](https://leetcode.com/problems/longest-common-prefix/)
- #### LATER Week 9 - Graph Resources
- LATER [Introduction to Graphs](https://youtu.be/gXgEDyodOJU)
- LATER [Multi-part Graph Lecture Series](https://youtu.be/c8P9kB1eun4)
- LATER [WilliamFiset's Graph Theory Videos](https://www.youtube.com/watch?v=DgXR2OWQnLc&list=PLDV1Zeh2NRsDGO4--qE8yH72HFL1Km93P)
- ### LATER Week 10 - Graphs
- LATER [Number of Connected Comps in Graph](https://protegejj.gitbook.io/algorithm-practice/leetcode/graph/323-number-of-connected-components-in-an-undirected-graph)
- LATER [Word Search II](https://leetcode.com/problems/word-search-ii/)
- LATER [Alien Dictionary](https://www.lintcode.com/problem/alien-dictionary/)
- LATER [Graph Valid Tree](https://www.lintcode.com/problem/graph-valid-tree/)
- LATER [Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)
- LATER [Find Median from Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)
- LATER [Daily Temp (**Bonus**)](https://leetcode.com/problems/daily-temperatures/)
- ### LATER Week 11 - Dynamic Programming
- LATER [Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)
- LATER [House Robber](https://leetcode.com/problems/house-robber/)
- LATER [House Robber II](https://leetcode.com/problems/house-robber-ii/)
- LATER [Coin Change](https://leetcode.com/problems/coin-change/)
- LATER [Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)
- LATER [BST Iterator (**Bonus**)](https://leetcode.com/problems/binary-search-tree-iterator/)
- #### LATER Week 11 - Dynamic Programming Resources
- LATER [What is Dynammic Programming and How to Use It](https://youtu.be/vYquumk4nWw) by CS Dojo
- LATER [5 Simple Steps for DP](https://youtu.be/aPQY__2H3tE) by Reducible
- LATER MIT Lectures on DP (numbered 19-21)
- LATER [Lecture 19](https://youtu.be/OQ5jsbhAv_M)
- LATER [Lecture 20](https://youtu.be/ENyox7kNKeY)
- LATER [Lecture 21](https://youtu.be/ocZMDMZwhCY)
- LATER [MIT Lectures on Advanced DP](https://youtu.be/Tw1k46ywN6E)
- ### LATER Week 12 - Dynamic Programming
- LATER [Unique Paths](https://leetcode.com/problems/unique-paths/)
- LATER [Jump Game](https://leetcode.com/problems/jump-game/)
- LATER [Decode Ways](https://leetcode.com/problems/decode-ways/)
- LATER [Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/)
- LATER [Word Break](https://leetcode.com/problems/word-break/)
- LATER [Combination Sum IV](https://leetcode.com/problems/combination-sum-iv/)
- LATER [Max Area of Island (**Bonus**)](https://leetcode.com/problems/max-area-of-island/)
- ### LATER Week 13 (Bonus) - Binary
- LATER [Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)
- LATER [Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)
- LATER [Counting Bits](https://leetcode.com/problems/counting-bits/)
- LATER [Missing Number](https://leetcode.com/problems/missing-number/)
- LATER [Reverse Bits](https://leetcode.com/problems/reverse-bits/)
- LATER [Developer-Tester Integration (**Bonus**)](https://leetcode.com/discuss/interview-question/882739/Snowflake-or-Summer-2021-Internship-OA-or-Developer-Tester-Integration)
- ### LATER Week 14 (Bonus) - Recursion and Backtracking
- LATER [Permutations](https://leetcode.com/problems/permutations/)
- LATER [Letter Combos of Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/)
- LATER [Sudoku Solver](https://leetcode.com/problems/sudoku-solver/)
- LATER [Subsets](https://leetcode.com/problems/subsets/)
- LATER [Generate Parentheses](https://leetcode.com/problems/generate-parentheses/)
- LATER [Palindrome Partitioning](https://leetcode.com/problems/palindrome-partitioning/)
- LATER [Flatten Nested List Iterator (**Bonus**)](https://leetcode.com/problems/flatten-nested-list-iterator/)
- ### LATER Week 15 (Bonus) - Binary Search
- LATER [Discussion](https://leetcode.com/discuss/general-discussion/786126/Python-Powerful-Ultimate-Binary-Search-Template.-Solved-many-problems)
- LATER [First Bad Version](https://leetcode.com/problems/first-bad-version/)
- LATER [Sqrt(x)](https://leetcode.com/problems/sqrtx/)
- LATER [Search Insert Position](https://leetcode.com/problems/search-insert-position/)
- LATER [Capacity to Ship Packages within D Days](https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/)
- LATER [Split Array Largest Sum](https://leetcode.com/problems/split-array-largest-sum/)
- LATER [Koko Eating Bananas](https://leetcode.com/problems/koko-eating-bananas/)
- LATER [Range Sum Query (Bonus)](https://leetcode.com/problems/range-sum-query-immutable/)
- LATER [Range Sum Query 2d Immutable (**Bonus**)](https://leetcode.com/problems/range-sum-query-2d-immutable/)

15
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- Todo:
- Personal
- DONE 订端传媒的新闻
SCHEDULED: <2023-04-01 Sat>
-
- Daily reflection [[Daily reflections]]
- What I've done
- 六点钟起床跑步，刷题
- 春游
- 演出
- What I've thought #thoughts
- 春游比较一般，在那个地方坐了很久，没任何规划
- 演出还行，鲁奕文气氛组但当
- Mood
- 开心

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- Leetcode Maximum Product Subarray
id:: 64261f45-6885-422c-808c-9344273af1d4
collapsed:: true
- Times:
- Completion time: 08:38
- Revisions:
- Tags:
- Algorithms: #Kadane_s_algorithm
- Data structures: #array
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/maximum-product-subarray/description/)
- Problem:
- Given an integer array `nums`, find a subarray that has the largest product, and return _the product_. The test cases are generated so that the answer will fit in a **32-bit** integer.
- Examples:
- ```
Example 1:
Input: nums = [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.
Example 2:
Input: nums = [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.
```
- Constraints:
- `1 <= nums.length <= 2 * 104`
- `-10 <= nums[i] <= 10`
- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.
- Thoughts:
- This is a variation of Kadane's algorithm:
- We don't change the `localMax` variable based on the value, but we only change it to 1 when it's 0, this is because multiplying a integer always give you a bigger abs. value.
- We want to have even number of negative numbers multiplied, when there are odd, one additional negative number must be avoided, to solve this problem, we iterate the array twice, from left to right and from right to left.
- Solution:
- ```java
class Solution {
public int maxProduct(int[] nums) {
// try kadane's algorithm
int max = 1;
int globalMax = Integer.MIN_VALUE;
int globalMax1 = Integer.MIN_VALUE;
for (int i : nums) {
max *= i;
globalMax = Math.max(max, globalMax);
if (max == 0) {
max = 1;
}
}
max = 1;
for (int i = nums.length - 1; i >= 0; i--) {
max *= nums[i];
globalMax1 = Math.max(max, globalMax1);
if (max ==0) {
max = 1;
}
}
return Math.max(globalMax, globalMax1);
}
}
```
-
- TODO:
- DONE 写一个 12 bar blues 音乐
SCHEDULED: <2023-04-03 Mon>

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- Todo
- 学习
- DONE 研究 F(x)和 P(X)的关系
SCHEDULED: <2023-04-02 Sun>

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- Leetcode Game of Life
id:: 6428db9f-8f14-41a4-b5ec-5e79ffeb204a
collapsed:: true
- Times:
- Time when completed: 10:19
- Time taken to complete: 27min
- Revisions:
- Tags:
- Data structures: #array
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/game-of-life/description/)
- Problem:
- According to [Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "The **Game of Life**, also known simply as **Life**, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."
The board is made up of an `m x n` grid of cells, where each cell has an initial state: **live** (represented by a `1`) or **dead** (represented by a `0`). Each cell interacts with its [eight neighbors](https://en.wikipedia.org/wiki/Moore_neighborhood) (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):
- Any live cell with fewer than two live neighbors dies as if caused by under-population.
- Any live cell with two or three live neighbors lives on to the next generation.
- Any live cell with more than three live neighbors dies, as if by over-population.
- Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.
The next state is created by applying the above rules
simultaneously to every cell in the current state, where births and
deaths occur simultaneously. Given the current state of the `m x n` grid `board`, return _the next state_.
- Follow up:
1. Could you solve it in-place? Remember that the board needs to be updated simultaneously: You cannot update some cells first and then use their updated values to update other cells.
2. In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches upon the border of the array (i.e., live cells reach the border). How would you address these problems?
- Examples:
- ```
Example 1:
Input: board = [[0,1,0],[0,0,1],[1,1,1],[0,0,0]]
Output: [[0,0,0],[1,0,1],[0,1,1],[0,1,0]]
Example 2:
Input: board = [[1,1],[1,0]]
Output: [[1,1],[1,1]]
```
- Constraints:
- `m == board.length`
- `n == board[i].length`
- `1 <= m, n <= 25`
- `board[i][j]` is `0` or `1`.
- Thoughts:
- This is a simple problem about iterating over 2d arrays, two difficulties here:
- to check for OOB
- Just write a function
- to do the steps in simultaneously
- Write count of neighbors to the boards first, but remember _to write 1 if the count is 0 and board value is 1_, otherwise it will be mistaken for 0 in follow up counting, and its gonna die anyways :)
- Solution:
- ```java
class Solution {
private boolean isLegit(int[][] board, int x, int y) {
// returns if a location is legit
if (x < 0 || y < 0) {
return false;
}
if (x >= board.length || y >= board[0].length) {
return false;
}
return true;
}
private int countNeighbors(int[][] board, int x, int y) {
// returns live neighbors of the cell
int count = 0;
for (int a = x - 1; a <= x + 1; a++) {
for (int b = y - 1; b <= y + 1; b++) {
if (!isLegit(board, a, b) || (a == x && b == y)) {
continue;
}
if (board[a][b] > 0) {
count++;
}
}
}
// System.out.println("Neighbors for " + x + " " + y + " is " + count);
return count;
}
public void gameOfLife(int[][] board) {
// update cells
int count;
for (int x = 0, lx = board.length; x < lx; x++) {
for (int y = 0, ly = board[0].length; y < ly; y++) {
count = countNeighbors(board, x, y);
if (board[x][y] == 1) {
if (count == 0) {
// going to die anyways
board[x][y] = 1;
} else {
board[x][y] = count;
}
} else {
board[x][y] = count * -1;
}
}
}
for (int x = 0, lx = board.length; x < lx; x++) {
for (int y = 0, ly = board[0].length; y < ly; y++) {
if (board[x][y] > 0) {
if (board[x][y] < 2) {
board[x][y] = 0;
} else if (board[x][y] > 3) {
board[x][y] = 0;
} else {
board[x][y] = 1;
}
} else if (board[x][y] == -3) {
board[x][y] = 1;
} else {
board[x][y] = 0;
}
}
}
}
}
```
-
- Leetcode Find Minimum in Rotated Sorted Array
id:: 64294296-85d5-4d21-b815-b4bad7525df5
collapsed:: true
- Times:
- Time when completed: 16:53
- Time taken to complete: 47min
- DONE Revisions:
SCHEDULED: <2023-05-05 Fri>
:LOGBOOK:
- State "DONE" from "LATER" [2023-04-21 Fri 14:38]
- State "DONE" from "LATER" [2023-04-21 Fri 14:39]
:END:
* [[Apr 21st, 2023]] 3min
- Tags:
- Algorithms: #binary_search
- Data structures: #array #interval
- Difficulty: #difficulty_medium
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/)
- Problem:
- Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:
- `[4,5,6,7,0,1,2]` if it was rotated `4` times.
- `[0,1,2,4,5,6,7]` if it was rotated `7` times.
- Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.
Given the sorted rotated array `nums` of **unique** elements, return _the minimum element of this array_.
You must write an algorithm that runs in `O(log n) time.`
- Examples:
- ```
Example 1:
Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:
Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.
```
- Constraints:
- `n == nums.length`
- `1 <= n <= 5000`
- `-5000 <= nums[i] <= 5000`
- All the integers of `nums` are **unique**.
- `nums` is sorted and rotated between `1` and `n` times.
- Thoughts:
- By analyzing the appearance of the array, we can know that there is a slope pattern:
```txt
4,5,1,2,3
-,t,l,-,-
-,o,o,-,-
-,p,w,-,-
```
We want to find the starting point of the slope, and we can make use of three data:
1. nums[l]
2. nums[r]
3. nums[mid]
When nums[mid] < nums[r], this means mid is on the right side slope, and we need to move r to mid.
When nums[mid] > nums[r], this means mid is on the left side slope, we need to move l to mid, and since we need the starting point of the right slope, we move l to mid + 1, and this breaks the loop when l == r - 1
- Solution:
- Revision 1: basic binary search method
```java
class Solution {
public int findMin(int[] nums) {
// Binary search for the change point.
int l = 0, r = nums.length - 1;
int mid = 0;
while (r > l) {
mid = l + (r - l) / 2;
if (nums[r] - nums[mid] > 0) {
// the change point is before mid;
r = mid;
} else if (nums[mid] - nums[l] >= 0) {
// after mid
l = mid + 1;
}
}
return nums[l];
}
}
```
- Revision 2: remove line 11's conditions
```java
class Solution {
public int findMin(int[] nums) {
// Binary search for the change point.
int l = 0, r = nums.length - 1;
int mid = 0;
while (r > l) {
mid = l + (r - l) / 2;
if (nums[r] - nums[mid] > 0) {
// the change point is before mid;
r = mid;
} else {
// after mid
l = mid + 1;
}
}
return nums[l];
}
}
```
-
- #+BEGIN_VERSE
毛概视频大纲 #to_be_deleted
视频内容：
按照时间顺序从古到今叙述，历史部分从建党百年的艰辛探索，到北邮校史中的艰苦奋斗和国家情怀，当代部分讲北邮的创新环境和青年人的强国梦想。以“强国有我”为线索，环环相扣地指出强国有我为什么行，当代青年为什么要奋斗共筑中国梦
特色：
采用实地取景的方式，图文并茂，身临其镜，致力于给观众带来最优质的学习体验
分工：
马闻良：演讲中国党史部分，并撰写有关稿件。
王子昂：演讲北邮校史部分，并撰写有关稿件。
路思远：演讲北邮当代环境，并撰写有关稿件。
李王子：演讲青年强国梦想，并撰写有关稿件。
黄源源：视频拍摄，后期剪辑、配乐、和字幕等编辑作业。
#+END_VERSE
-
- Todo
- CANCELED [#B] 看电影
- DONE [#B] 整理书单和电影清单
SCHEDULED: <2023-04-04 Tue>
- DONE 写视频大纲
:LOGBOOK:
CLOCK: [2023-04-02 Sun 19:29:35]--[2023-04-02 Sun 19:49:05] => 00:19:30
:END:
- DONE 写一个每周 todo 的模板
SCHEDULED: <2023-04-03 Mon>
-
- Leetcode Search in Rotated Sorted Array
id:: 642aa126-13d8-48ff-8eb4-f17d875add97
collapsed:: true
- Times:
- Time when completed: 17:41
- Time taken to complete: 26min
- DONE Revisions:
SCHEDULED: <2023-05-05 Fri>
:LOGBOOK:
- State "DONE" from "LATER" [2023-04-21 Fri 15:14]
:END:
* [[Apr 21st, 2023]] 9min
- Tags:
- Algorithms: #binary_search
- Data structures: #array #interval
- Difficulty: #difficulty_medium
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/search-in-rotated-sorted-array/description/)
- Problem:
- There is an integer array `nums` sorted in ascending order (with **distinct** values).
- Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index `3` and become `[4,5,6,7,0,1,2]`.
- Given the array `nums` **after** the possible rotation and an integer `target`, return _the index of _`target`_ if it is in _`nums`_, or _`-1`_ if it is not in _`nums`.
- You must write an algorithm with `O(log n)` runtime complexity.
- Examples:
- ```
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
Example 3:
Input: nums = [1], target = 0
Output: -1
```
- Constraints:
- `1 <= nums.length <= 5000`
- `-104 <= nums[i] <= 104`
- All values of `nums` are **unique**.
- `nums` is an ascending array that is possibly rotated.
- `-104 <= target <= 104`
- Thoughts:
- This is a binary search problem, should take some more time to understand it, also, see ((64294296-85d5-4d21-b815-b4bad7525df5))
- Solution:
- ```java
class Solution {
int biSearch(int[] nums, int l, int r, int target) {
// System.out.printf("l: %d, r: %d\n", l, r);
if (l > r || target < nums[l] || target > nums[r]) {
return -1;
}
int mid;
while (l <= r) {
mid = l + (r - l) / 2;
if (nums[mid] > target) {
r = mid - 1;
} else if (nums[mid] < target) {
l = mid + 1;
} else {
return mid;
}
}
return -1;
}
public int search(int[] nums, int target) {
// first, find the pivot index, second, use binary search
int l = 0, r = nums.length - 1, mid;
while (l < r) {
mid = l + (r - l) / 2; // check notes for why
if (nums[r] > nums[mid]) {
r = mid;
} else {
l = mid + 1;
}
}
final int starting = l;
// System.out.println(starting);
int ansl = biSearch(nums, 0, starting - 1, target);
int ansr = biSearch(nums, starting, nums.length - 1, target);
return Math.max(ansl, ansr);
}
}
```
-
- [[GPT notes]]
collapsed:: true
- How to write a 12 bar jazz? #music
- To write a 12-bar blues, first, choose a key for your song. The traditional blues progression is based on the I-IV-V chord progression. In the key of C, this would be C, F, and G.
Next, write the lyrics. A 12-bar blues typically follows an AAB lyrical structure, with the A line being repeated and the B line concluding the phrase.
The 12-bar structure is as follows:
```text
| I | I | I | I | IV | IV | I | I | V | IV | I | I |
```
In jazz, the 12-bar structure can be similar to the blues, but with more complex chords and often a different arrangement. Jazz musicians often use ii-V-I progressions to add more color and complexity to the harmony. It's important to experiment with different chord progressions and song structures to create a unique and interesting jazz composition.
- How to learn to write music? #music
- Learning to write music is a creative process that requires patience, practice, and dedication. Here are some steps that can help you get started:
- 1. Study music theory: Understanding music theory is crucial to write good music. You should start by learning the basics of rhythm, melody, and harmony.
- 2. Listen to music: Pay attention to the music you enjoy and try to analyze it. Listen to the different parts of the instrumentation and take notice of how the different elements come together to create a cohesive melody.
- 3. Experiment with different instruments: If you’re a beginner, experiment with different instruments to find the one that best fits you. You may also try using music software to create music.
- 4. Write a melody: One of the easiest ways to start writing music is to come up with a melody. Once you have a melody in mind, experiment with different chords and harmonies to see what works best.
- 5. Practice regularly: The key to getting better at writing music is to practice regularly. Set aside some time each day to create and experiment with different sounds.
- 6. Get feedback: Listen to what others have to say about your music. Take constructive criticism from experienced musicians and work on improving your compositions accordingly.
- Remember that everyone has their unique style when it comes to writing music. Don't be afraid to experiment and try new things. Keep practicing, and you will eventually find your rhythm.

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- Todo
- DONE 洗衣服
- DONE 做数据库 lab
:LOGBOOK:
CLOCK: [2023-04-03 Mon 11:21:56]--[2023-04-03 Mon 12:18:36] => 00:56:40
CLOCK: [2023-04-03 Mon 12:48:48]--[2023-04-03 Mon 12:48:49] => 00:00:01
CLOCK: [2023-04-03 Mon 16:48:53]--[2023-04-03 Mon 17:48:37] => 00:59:44
:END:
- DONE Figure out how to manage use flags for gentoo ( by using eix)
:LOGBOOK:
CLOCK: [2023-04-03 Mon 13:27:52]--[2023-04-03 Mon 13:27:54] => 00:00:02
:END:
- LATER 复习概率论
collapsed:: true
- LATER 公式法
- LATER 求期望
:LOGBOOK:
CLOCK: [2023-04-03 Mon 14:25:17]
:END:
- LATER 积分的公式
- DONE [#A] 写学交论文
SCHEDULED: <2023-04-11 Tue .+1d>
DEADLINE: <2023-04-09 Sun>
:LOGBOOK:
CLOCK: [2023-04-04 Tue 13:56:38]--[2023-04-04 Tue 16:03:06] => 02:06:28
CLOCK: [2023-04-07 Fri 13:35:02]--[2023-04-07 Fri 13:46:01] => 00:10:59
- State "DONE" from "NOW" [2023-04-07 Fri 13:46]
* State "DONE" from "LATER" [2023-04-09 Sun 16:57]
CLOCK: [2023-04-09 Sun 16:57:28]--[2023-04-09 Sun 16:57:29] => 00:00:01
:END:
* DONE 内容
- DONE 添加边缘计算对碳中和的影响
- AI 对论文修改的建议： [[GPT notes]]
#+BEGIN_VERSE
To insert the new section on "Carbon Neutrality and Edge Computing" into the paper, you would need to determine the appropriate location based on the content of the section. Generally, this section could be inserted after the section on "The Benefits of Edge Computing," as it is related to the topic of sustainability and energy efficiency in edge computing.
Here is an updated outline of the paper with the new section added:
I. Introduction
A. Definition of Edge Computing
B. Importance of Edge Computing
C. Purpose of the Paper
II. How Edge Computing Works
A. Basic Architecture of Edge Computing
B. Characteristics of Edge Computing
C. Comparison with Cloud Computing
III. Applications of Edge Computing
A. IoT and Smart Devices
B. Autonomous Vehicles
C. Healthcare
D. Industry 4.0
E. Smart Grid
IV. The Benefits of Edge Computing
A. Reduced Latency and Improved Response Time
B. Increased Security and Privacy
C. Reduced Bandwidth and Connectivity Costs
D. Improved Reliability and Availability
E. Enhanced User Experience
F. Carbon Neutrality and Energy Efficiency
V. Carbon Neutrality and Edge Computing
A. Definition of Carbon Neutrality
B. Importance of Carbon Neutrality in Edge Computing
C. Techniques for Achieving Carbon Neutrality in Edge Computing
D. Impact of Carbon Neutrality on Business Operations
VI. Challenges and Limitations of Edge Computing
A. Security Risks
B. Integration with Legacy Systems
C. Limited Computing Resources
D. Complexity of Deployment and Management
VII. Future of Edge Computing
A. Trends and Innovations
B. Potential Applications
C. Impact on Society
VIII. Conclusion
A. Summary of Key Points
B. Implications for Businesses and Society
C. Recommendations for Further Research.
#+END_VERSE
- DONE 用 AI 修改摘要
- DONE 加字数 300 字
- DONE 加引用 2 个 \* LATER 格式：看 IEEE 的格式
- DONE 概率论论文
DEADLINE: <2023-04-12 Wed>
- DONE 参加商谈总决赛
SCHEDULED: <2023-04-08 Sat>
- DONE 给杨哥机器学习的代码
- DONE 护眼程序：提醒我休息眼睛 #projects
- DONE edx 音乐课最终作业
DEADLINE: <2023-04-07 Fri>
-
- Daily reflection [[Daily reflections]]
- What I've done
- 看书
- 学 leetcode
- 做一堆 todo
- What I've thought #thoughts
- 看了不少书，这挺好，要坚持下去
- DONE 写了很多 leetcode，但是很多问题都是直接背答案，不知道为什么要用那个方法，应该看下算法书。
- Mood
- good
-
- [[New computer 2023]]
- Try the new pkgcore from gentoo
-
- Leetcode 3Sum
id:: 642abd85-5c14-42c3-aa39-e7dc2773b2ec
- Times:
- Time when completed: 19:50
- Time taken to complete: 47min
- DONE Revisions: do it again in c++
SCHEDULED: <2023-05-19 Fri>
:LOGBOOK:
- State "DONE" from "LATER" [2023-04-21 Fri 16:05]
* State "DONE" from "LATER" [2023-05-23 Tue 20:26]
:END:
* [[Apr 21st, 2023]] 12mins java
- Tags:
- Algorithms: #two_pointers
- Data structures: #array #set #interval
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/3sum/description/)
- Problem:
- Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.
- Notice that the solution set must not contain duplicate triplets.
- Examples:
- ```
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
```
- Constraints:
- `3 <= nums.length <= 3000`
- `-105 <= nums[i] <= 105`
- Thoughts:
- #+BEGIN_VERSE
This is a O(n^2) two pointer problem.
First, we use a variable `i`, then between i + 1 and the end, we use two pointers to find the solution that adds up to 0, and store it in a set (it will not be duplicated)
The main difficulty here is to get to know java.
#+END_VERSE
- Solution:
- ```java
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
// modified two pointer method
Arrays.sort(nums);
Set<List<Integer>> ans = new HashSet();
for (int i = 0; i < nums.length - 2; i++) {
int l = i + 1;
int r = nums.length - 1;
while (l < r) {
int sum = nums[i] + nums[l] + nums[r];
if (sum > 0) {
r--;
} else if (sum < 0) {
l++;
} else {
ans.add(Arrays.asList(nums[l], nums[r], nums[i])); //java's set will check for dupes by default.
l++;
r--;
}
}
}
return new ArrayList<List<Integer>>(ans);
}
}
```
-
- Leetcode Container With Most Water
id:: 642ac7bd-a908-4155-b839-f58ca64f4672
- Times:
- Time when completed: 20:34
- Time taken to complete: 30min
- DONE Revisions:
SCHEDULED: <2023-05-19 Fri>
:LOGBOOK:
- State "DONE" from "LATER" [2023-04-21 Fri 15:23]
:END:
* [[Apr 21st, 2023]] 10min
- Tags:
- Algorithms: #greedy #two_pointers
- Data structures: #array #interval
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/container-with-most-water/description/)
- Problem:
- You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`.
- Find two lines that together with the x-axis form a container, such that the container contains the most water.
- Return _the maximum amount of water a container can store_.
- **Notice** that you may not slant the container.
- Examples:
- ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
- ```
Example 1:
Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example 2:
Input: height = [1,1]
Output: 1
```
- Constraints:
- `n == height.length`
- `2 <= n <= 10^5`
- `0 <= height[i] <= 104`
- Thoughts:
- There are 10^5 numbers in the array, so the O(N^2) brute force is not an option.
- By analyzing this problem, we can use greedy algorithm, because the only way we can get a better result is by move the shortest line in two sides in a longer one, and we simply keep track of this interval using two pointers method.
- 1. For the first iteration of our answer, we can keep track of the two sides using two pointers `l` and `r`, and in a loop we decide which one to increment / reduce by calculating which one's height is bigger, and we move the side which has a smaller height.
- 2. To optimize the answer we can think about the algorithm in step one: _When we are moving one side, the result might be **even smaller** if t**he new height is smaller** than the original one_. To address this issue, we can add a while loop and a marker variable `smol` to keep track of the smaller side.
- Solution:
- ```java
class Solution {
public int maxArea(int[] height) {
// Use greedy algorithm with two pointers
int l = 0, r = height.length - 1;
int max = 0;
while (l < r) {
max = Math.max(max, (r - l) * Math.min(height[l], height[r]));
// the sum will only be bigger if the next line is higher.
int smol = Math.min(height[l], height[r]);
// only one of the loop will run
while (height[l] <= smol && l < r) {
l++;
}
while (height[r] <= smol && l < r) {
r--;
}
}
return max;
}
}
```

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- Todo
- DONE Clean software in my container
:LOGBOOK:
CLOCK: [2023-04-12 Wed 20:14:17]--[2023-04-13 Thu 09:13:14] => 12:58:57
:END:
- LATER 整理游戏库
-
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- What I've thought #thoughts
collapsed:: true
- Dreamed of being kicked out from japanese class for bad at grades, and have to stick with the 60 points.
- Mood

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- Leetcode Insert Interval
id:: 642cd1a8-9707-4dc2-9d03-11464553e618
- Times:
- Time when completed: 10:05
- Time taken to complete: 20min (I looked up solution from chatGPT)
- Revisions:
- Tags:
- Algorithms: #greedy
- Data structures: #array #interval
- Difficulty: #difficulty_medium
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/insert-interval/description/)
- Problem:
- You are given an array of non-overlapping intervals `intervals` where `intervals[i] = [starti, endi]` represent the start and the end of the `ith` interval and `intervals` is sorted in ascending order by `starti`. You are also given an interval `newInterval = [start, end]` that represents the start and end of another interval.
- Insert `newInterval` into `intervals` such that `intervals` is still sorted in ascending order by `starti` and `intervals` still does not have any overlapping intervals (merge overlapping intervals if necessary).
- Return `intervals`_ after the insertion_.
- Examples:
- ```
Example 1:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
Example 2:
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
```
- Constraints:
- `0 <= intervals.length <= 104`
- `intervals[i].length == 2`
- `0 <= starti <= endi <= 105`
- `intervals` is sorted by `starti` in **ascending** order.
- `newInterval.length == 2`
- `0 <= start <= end <= 105`
- Thoughts:
- AI's solution:
collapsed:: true
- ```java
class Solution {
public int[][] insert(int[][] intervals, int[] newInterval) {
List<int[]> result = new ArrayList<>();
int i = 0;
// add intervals that end before newInterval starts
while (i < intervals.length && intervals[i][1] < newInterval[0]) {
result.add(intervals[i]);
i++;
}
// merge overlapping intervals with newInterval
while (i < intervals.length && intervals[i][0] <= newInterval[1]) {
newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
i++;
}
result.add(newInterval);
// add remaining intervals
while (i < intervals.length) {
result.add(intervals[i]);
i++;
}
return result.toArray(new int[result.size()][2]);
}
}
```
- The solution uses a greedy approach to merge the given intervals with the new interval. We first add all the intervals that end before the new interval starts, as they cannot overlap with the new interval. Then we merge any overlapping intervals with the new interval by updating its start and end values. Finally, we add any remaining intervals that start after the new interval ends.
- The time complexity of this solution is O(n), where n is the number of intervals in the input array. This is because we iterate over the intervals array only once. The space complexity is also O(n), as we may need to store all the intervals in the result array.
- Intuition:
- We use greedy algorithm here, since we are going to find the optimal
solution at each step. The code itself is simple to read and understand,
the difficulty for me to write the code is to check when intervals
overlap and how to merge them.
- Approach:
- The code has four steps:
- Construct an answer arrayList, and add every interval before `newInterval` to the answer
- Use a while loop, to merge overlapping intervals into `newIntreval`, and later we skip these already merged interval
- Add the `newInterval` to the answer array
- Add remaining intervals to the answer
- Solution:
- The code is $$O(n)$$ time $$O(n)$$ space
```java
class Solution {
public int[][] insert(int[][] intervals, int[] newInterval) {
// Simulation doesn't work lol
// We just merge the intervals
ArrayList<int[]> ans = new ArrayList();
// add the intervals before newInterval
int i = 0;
while (i < intervals.length && newInterval[0] > intervals[i][1]) {
ans.add(intervals[i]);
i++;
}
// merge all overlapping interval into newInterval
while (i < intervals.length && newInterval[1] >= intervals[i][0]) {
newInterval[0] = Math.min(intervals[i][0], newInterval[0]);
newInterval[1] = Math.max(intervals[i][1], newInterval[1]);
i++;
}
ans.add(newInterval);
// add the remaining to answer
while (i < intervals.length) {
ans.add(intervals[i++]);
}
return ans.toArray(new int[ans.size()][2]);
}
}
```
-
- Leetcode - Length of Last Word
- Times:
- Time when completed: 10:49
- Time taken to complete: about 5 mins
- Revisions:
- Tags:
- Data structures: #array
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/length-of-last-word/description/)
- Problem:
- Given a string `s` consisting of words and spaces, return _the length of the **last** word in the string._
- A **word** is a maximal substring consisting of non-space characters only.
- Examples:
- ```
Example 1:
Input: s = "Hello World"
Output: 5
Explanation: The last word is "World" with length 5.
Example 2:
Input: s = " fly me to the moon "
Output: 4
Explanation: The last word is "moon" with length 4.
Example 3:
Input: s = "luffy is still joyboy"
Output: 6
Explanation: The last word is "joyboy" with length 6.
```
- Constraints:
- `1 <= s.length <= 104`
- `s` consists of only English letters and spaces `' '`.
- There will be at least one word in `s`.
- Thoughts:
- Intuition:
- The most important part is to count from end, this avoids traversing the entire array and helps with speed. $$O(n)$$ time and space.
- Approach:
- First, skip any spaces
- Then start counting until a space is encountered
- Solution:
- Code
```java
class Solution {
public int lengthOfLastWord(String s) {
int len = 0;
int i = s.length() - 1;
while (i >= 0 && s.charAt(i) == ' ') {
i--;
}
while (i >= 0 && s.charAt(i) != ' ') {
i--;
len++;
}
return len;
}
}
```

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- Leetcode - Merge Intervals
id:: 642edfd2-c458-484e-9e94-8802771475d5
collapsed:: true
- Times:
- Time when completed: 09:34
- Time taken to complete: 37mins
- DONE Revisions:
SCHEDULED: <2023-05-19 Fri>
:LOGBOOK:
- State "DONE" from "LATER" [2023-04-21 Fri 15:47]
:END:
* [[Apr 21st, 2023]] 10mins
- Tags:
- Algorithms:
- Data structures: #array #interval
- Difficulty: #difficulty_medium
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/merge-intervals/description/)
- Problem:
- Given an array of `intervals` where `intervals[i] = [starti, endi]`, merge all overlapping intervals, and return _an array of the non-overlapping intervals that cover all the intervals in the input_.
- Examples:
- ```
Example 1:
Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
Example 2:
Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
```
- Constraints:
- `1 <= intervals.length <= 104`
- `intervals[i].length == 2`
- `0 <= starti <= endi <= 104`
- Thoughts:
- My take:
- Intuition:
- Similar to ((642cd1a8-9707-4dc2-9d03-11464553e618)), to merge all intervals in O(n), we have to sort the intervals array first, which takes $$O(nlogn)$$, and then we use the merging techniques similar to aforementioned solution.
- Approach:
- Sort the intervals
- use a `merged` to keep track current array, and use `ans` to keep track of the answer
- loop over the intervals, merge any intersecting intervals into merged
- when no intersections are found, push `merged` into `ans`
- return the array version of `ans`: `ans.toArray(..)`
- why we sort by their start times instead of end times like in [[Leetcode Non-Overlapping-Intervals]] ?
- We sort the intervals by their start times because it allows us to merge overlapping intervals more efficiently.
- Consider the case where we sort the intervals by their end times instead of start times. If we iterate over the sorted intervals and try to merge overlapping intervals by comparing their end times, we may miss some overlapping intervals.
- For example, suppose we have the intervals [1, 4], [2, 3], and [3, 5]. If we sort these intervals by their end times, we get [2, 3], [1, 4], and [3, 5]. If we iterate over these intervals and try to merge overlapping intervals by comparing their end times, we would merge [2, 3] and [1, 4] into [1, 4], but we would not merge [1, 4] and [3, 5], because their end times do not overlap.
- On the other hand, if we sort the intervals by their start times, we can be sure that if two intervals overlap, their start times will also overlap. This means that we can simply compare the end time of the previous interval with the start time of the current interval to determine whether they overlap, and we don't have to worry about missing any overlapping intervals.
- In summary, sorting the intervals by their start times allows us to merge overlapping intervals more efficiently and reliably.
- Solution:
- Code
- ```java
class Solution {
public int[][] merge(int[][] intervals) {
ArrayList<int[]> ans = new ArrayList();
Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
int[] merged = intervals[0];
int i = 1;
while (i < intervals.length) {
if (merged[1] >= intervals[i][0]) {
// they must overlap, because intervals are sorted.
merged[1] = Math.max(intervals[i][1], merged[1]);
} else {
ans.add(merged);
merged = intervals[i];
}
i++;
}
// there will always be a merged left not added.
ans.add(merged);
return ans.toArray(new int[ans.size()][2]);
}
}
```
-
- Leetcode - Palindrome Number
collapsed:: true
- Times:
- Time when completed: 13:55
- Time taken to complete: 3min
- Revisions:
- Tags:
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/palindrome-number/description/)
- Problem:
- Given an integer `x`, return `true` if `x` is a _palindrome_, and `false` otherwise
- Examples:
- ```
Example 1:
Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.
Example 2:
Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
```
- Constraints:
- `-2^31 <= x <= 2^31 - 1`
- Thoughts:
- Intuition:
- If a number is a palindrome, the reverse of it equals its original. We reverse a number and check if they are equal
- Special cases should be considered:
- 0
- negative
- has 0 on the end
- Approach:
- Reverse a number
- check if they equals
- Solution:
- Code
- ```java
class Solution {
public boolean isPalindrome(int x) {
if (x == 0) {
return true;
}
if (x < 0 || x % 10 == 0) {
return false;
}
// make an reverse number
int reverse = 0;
int copy = x;
while (copy > 0) {
reverse = reverse * 10 + copy % 10;
copy /= 10;
}
return reverse == x;
}
}
```
-
- DONE 把智慧树剩下的一点课刷了
SCHEDULED: <2023-04-06 Thu>
-
- DONE 规划乐理学习
- Open Music Theory 的记录
- 日常 ear training
-
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- Music theory course
- Future plans for music theory study
- Spent a lot of time improvising
- properly configured ardour and stuff
- What I've thought #thoughts
- Music theory is fun, I want to freely express myself one day.
- I figured reasons for not learning bass in school, I should learn more music theory instead.
- Mood

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- DONE Ardour
- 怎么录制
- Ardour vs. audacity
-
- Leetcode - Happy Number
collapsed:: true
- Times:
- Time when completed: 12:36
- Time taken to complete: I forgot, very quick I suppose
- Revisions:
- Tags:
- Algorithms: #Floyd_s_cycle_finding_algorithm
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/happy-number/description/)
- Problem:
- Write an algorithm to determine if a number `n` is happy.
A **happy number** is a number defined by the following process:
- Starting with any positive integer, replace the number by the sum of the squares of its digits.
- Repeat the process until the number equals 1 (where it will stay), or it **loops endlessly in a cycle** which does not include 1.
- Those numbers for which this process **ends in 1** are happy.
Return `true` _if_ `n` _is a happy number, and_ `false` _if not_.
- Examples:
- ```
Example 1:
Input: n = 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
Example 2:
Input: n = 2
Output: false
```
- Constraints:
- `1 <= n <= 231 - 1`
- Thoughts:
- Intuition:
- It loops endlessly in a cycle, which means we need to use a cycle detection algorithm, which happens to be the [[Floyd's Cycle Finding Algorithm]].
- Approach:
- Use the Floyd's cycle finding algorithm, with two variables
- when fast == slow, cycle is found
- detect if fast is 1, and return values
- Solution:
- Code
- ```java
class Solution {
private int progress(int n) {
int sum = 0;
while (n > 0) {
sum += (n % 10) * (n % 10);
n /= 10;
}
return sum;
}
public boolean isHappy(int n) {
// loop detection
int slow = progress(n);
int fast = progress(slow);
while (slow != fast) {
slow = progress(slow);
fast = progress(progress(fast));
}
return (fast == 1);
}
}
```
-
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- 写学交论文
- 音乐 improvising
- I met someone, she's gorgeous
- What I've thought #thoughts
- Improvisation is fun :)
- I think I fall in love with someone, I have a strong feeling when she is nearby, and dopamine burst out whenever I talk to her. I don't know if she feel the same (I think to some extent she does. I don't really know, I just follow the flow and enjoy every moment of that). ~~aaand also I erected a lot~~
- Mood
- Generally excited, but sometimes exhausted from these happy feelings, I deserve a good sleep.
-
- Todo
- DONE Edx 写一篇 blues
:LOGBOOK:
CLOCK: [2023-04-07 Fri 14:36:18]--[2023-04-07 Fri 16:50:36] => 02:14:18
:END:
- DONE 概率论作业
:LOGBOOK:
CLOCK: [2023-04-07 Fri 17:31:14]--[2023-04-07 Fri 21:19:06] => 03:47:52
:END:
- DONE 22, 23, 24, 25, 26, 27, 29
- DONE 58, 59
-
- 乐理学习 #music
- LATER 每天学 15 分钟 Open Music Theory [上次学到的地方](https://viva.pressbooks.pub/openmusictheory/chapter/aspn/)
SCHEDULED: <2023-05-16 Tue .+1d>
:LOGBOOK:
- State "DONE" from "LATER" [2023-04-07 Fri 16:50]
- State "DONE" from "LATER" [2023-04-07 Fri 16:52]
* State "DONE" from "LATER" [2023-04-10 Mon 22:49]
* State "DONE" from "LATER" [2023-04-11 Tue 23:35]
CLOCK: [2023-04-12 Wed 09:16:56]--[2023-04-12 Wed 09:44:02] => 00:27:06
* State "DONE" from "LATER" [2023-04-12 Wed 09:44]
* State "DONE" from "LATER" [2023-04-18 Tue 23:31]
* State "DONE" from "LATER" [2023-04-21 Fri 09:30]
CLOCK: [2023-04-23 Sun 15:18:42]--[2023-04-23 Sun 15:18:43] => 00:00:01
* State "DONE" from "NOW" [2023-04-23 Sun 15:18]
* State "DONE" from "LATER" [2023-04-26 Wed 22:31]
* State "DONE" from "LATER" [2023-05-08 Mon 20:47]
CLOCK: [2023-05-09 Tue 09:52:58]--[2023-05-09 Tue 10:01:36] => 00:08:38
* State "DONE" from "NOW" [2023-05-09 Tue 10:01]
CLOCK: [2023-05-09 Tue 10:11:30]--[2023-05-09 Tue 10:11:31] => 00:00:01
CLOCK: [2023-05-11 Thu 10:14:15]--[2023-05-11 Thu 11:16:14] => 01:01:59
* State "DONE" from "NOW" [2023-05-11 Thu 11:16]
* State "DONE" from "LATER" [2023-05-15 Mon 16:23]
:END:
- LATER 每天 5 分钟 Ear training [Teoria](https://www.teoria.com/en/exercises/)
SCHEDULED: <2023-05-12 Fri .+1d>
:LOGBOOK:
- State "DONE" from "LATER" [2023-04-07 Fri 23:31]
* State "DONE" from "LATER" [2023-04-11 Tue 16:41]
* State "DONE" from "LATER" [2023-04-12 Wed 15:07]
CLOCK: [2023-04-21 Fri 14:41:07]--[2023-04-21 Fri 14:41:09] => 00:00:02
* State "DONE" from "LATER" [2023-04-25 Tue 11:33]
* State "DONE" from "LATER" [2023-04-26 Wed 22:31]
* State "DONE" from "LATER" [2023-05-08 Mon 20:48]
CLOCK: [2023-05-09 Tue 10:01:48]--[2023-05-09 Tue 10:10:01] => 00:08:13
* State "DONE" from "LATER" [2023-05-09 Tue 10:10]
* State "DONE" from "LATER" [2023-05-11 Thu 20:06]
:END:

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- Daily reflection [[Daily reflections]]
- What I've done
collapsed:: true
- 昨天早上 4 点半醒，然后 6 点晨跑，白天军训
- 昨天晚上在商谈队参加共赢杯，拍照片
- 之后去吃海底捞吃“夜宵”
- 吃完夜宵还不尽兴，去唱 ktv
- 在 ktv 觥筹交错，凌晨两点左右鲁奕文和石皓钦先后倒下，我和于笑宇举杯畅饮，高歌 n 曲
- 点的歌都唱完了，正准备走的时候我说想要去看日出，然后拉上于和另外两人
- 经过热烈讨论决定去天安门
- 到了天安门之后发现我没有预约，鲁打算去什刹海，我赞同，下车后发现凌晨的北京巨冷，四下找旅店
- 旅馆奸商让我们四人睡一张床，在床上复盘，发现我们很智障，鲁碰我哪里我不记得了，我巨想睡觉，一个小时后睡过了日出
- 石坚持回去，鲁想再睡然后去玩，经过一番拉扯后还是都回了学校
- What I've thought #thoughts
collapsed:: true
- 经过我们的复盘，得出一个结论：喝了酒之后不要做决定
- 我之前有点难过，因为我觉得我不会在中国结婚，可能会因此影响到感情的进程，但后来又想，因为这个就 turn back 她显然更不负责任；而且我还是对双方有信心的，就算真的只有两三年的时间，我相信我们也还是能够促进互相进步，深入对自身的了解，抑或是创造回忆，这样即使不能结婚之类的对双方也是大有裨益的；再者，到那个时候还很早，现在也不该想太多
今天没学音乐，有点可惜
- Mood
collapsed:: true
- 有一点小小地抑郁，因为上述内容
- ecstatic
-
- 今日 Todo
- DONE 导出照片
:LOGBOOK:
CLOCK: [2023-04-09 Sun 19:21:18]--[2023-04-09 Sun 19:33:34] => 00:12:16
:END:
- DONE 学数据库
:LOGBOOK:
CLOCK: [2023-04-09 Sun 19:36:07]--[2023-04-09 Sun 21:41:50] => 02:05:43
:END:
- DONE leetcode
- DONE 比分
:LOGBOOK:
CLOCK: [2023-04-09 Sun 19:21:22]--[2023-04-09 Sun 19:26:41] => 00:05:19
:END:

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- Leetcode SQL Study Plan I #leetcode #excluded_in_query
id:: 6433ff05-fb6e-4a21-9bef-37b8fdcff801
collapsed:: true
- # Day 1
- DONE [595. Big Countries](https://leetcode.com/problems/big-countries/?envType=study-plan&id=sql-i)
:LOGBOOK:
CLOCK: [2023-04-10 Mon 20:25:30]--[2023-04-10 Mon 20:27:46] => 00:02:16
CLOCK: [2023-04-10 Mon 20:27:46]--[2023-04-10 Mon 20:27:47] => 00:00:01
:END:
- DONE [1757. Recyclable and Low Fat Products](https://leetcode.com/problems/recyclable-and-low-fat-products/?envType=study-plan&id=sql-i)
:LOGBOOK:
CLOCK: [2023-04-10 Mon 20:19:15]
CLOCK: [2023-04-10 Mon 20:28:52]--[2023-04-10 Mon 20:31:30] => 00:02:38
:END:
- DONE [584. Find Customer Referee](https://leetcode.com/problems/find-customer-referee/?envType=study-plan&id=sql-i)
:LOGBOOK:
CLOCK: [2023-04-10 Mon 20:19:22]--[2023-04-10 Mon 20:19:23] => 00:00:01
:END:
- DONE [183. Customers Who Never Order](https://leetcode.com/problems/customers-who-never-order/?envType=study-plan&id=sql-i)
- # Day 2
- DONE [1873. Calculate Special Bonus](https://leetcode.com/problems/calculate-special-bonus/?envType=study-plan&id=sql-i)
- LATER [627. Swap Salary](https://leetcode.com/problems/swap-salary/?envType=study-plan&id=sql-i)
- LATER [196. Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/?envType=study-plan&id=sql-i)
- # Day 3
- LATER [1667. Fix Names in a Table](https://leetcode.com/problems/fix-names-in-a-table/?envType=study-plan&id=sql-i)
- LATER [1484. Group Sold Products By The Date](https://leetcode.com/problems/group-sold-products-by-the-date/?envType=study-plan&id=sql-i)
- LATER [1527. Patients With a Condition](https://leetcode.com/problems/patients-with-a-condition/?envType=study-plan&id=sql-i)
- # Day 4
- LATER [1965. Employees With Missing Information](https://leetcode.com/problems/employees-with-missing-information/?envType=study-plan&id=sql-i)
- LATER [1795. Rearrange Products Table](https://leetcode.com/problems/rearrange-products-table/?envType=study-plan&id=sql-i)
- LATER [608. Tree Node](https://leetcode.com/problems/tree-node/?envType=study-plan&id=sql-i)
- LATER [176. Second Highest Salary](https://leetcode.com/problems/second-highest-salary/?envType=study-plan&id=sql-i)
- # Day 5
- LATER [175. Combine Two Tables](https://leetcode.com/problems/combine-two-tables/?envType=study-plan&id=sql-i)
:LOGBOOK:
CLOCK: [2023-04-10 Mon 20:21:45]--[2023-04-10 Mon 20:21:46] => 00:00:01
:END:
- LATER [1581. Customer Who Visited but Did Not Make Any Transactions](https://leetcode.com/problems/customer-who-visited-but-did-not-make-any-transactions/?envType=study-plan&id=sql-i)
- LATER [1148. Article Views I](https://leetcode.com/problems/article-views-i/?envType=study-plan&id=sql-i)
- # Day 6
- LATER [197. Rising Temperature](https://leetcode.com/problems/rising-temperature/?envType=study-plan&id=sql-i)
- LATER [607. Sales Person](https://leetcode.com/problems/sales-person/?envType=study-plan&id=sql-i)
- # Day 7
- LATER [1141. User Activity for the Past 30 Days I](https://leetcode.com/problems/user-activity-for-the-past-30-days-i/?envType=study-plan&id=sql-i)
- LATER [1693. Daily Leads and Partners](https://leetcode.com/problems/daily-leads-and-partners/?envType=study-plan&id=sql-i)
- LATER [1729. Find Followers Count](https://leetcode.com/problems/find-followers-count/?envType=study-plan&id=sql-i)
- # Day 8
- LATER [586. Customer Placing the Largest Number of Orders](https://leetcode.com/problems/customer-placing-the-largest-number-of-orders/?envType=study-plan&id=sql-i)
- LATER [511. Game Play Analysis I](https://leetcode.com/problems/game-play-analysis-i/?envType=study-plan&id=sql-i)
- LATER [1890. The Latest Login in 2020](https://leetcode.com/problems/the-latest-login-in-2020/?envType=study-plan&id=sql-i)
- LATER [1741. Find Total Time Spent by Each Employee](https://leetcode.com/problems/find-total-time-spent-by-each-employee/?envType=study-plan&id=sql-i)
- # Day 9
- LATER [1393. Capital Gain/Loss](https://leetcode.com/problems/capital-gainloss/?envType=study-plan&id=sql-i)
- LATER [1407. Top Travellers](https://leetcode.com/problems/top-travellers/?envType=study-plan&id=sql-i)
- LATER [1158. Market Analysis I](https://leetcode.com/problems/market-analysis-i/?envType=study-plan&id=sql-i)
- # Day 10
- LATER [182. Duplicate Emails](https://leetcode.com/problems/duplicate-emails/?envType=study-plan&id=sql-i)
- LATER [1050. Actors and Directors Who Cooperated At Least Three Times](https://leetcode.com/problems/actors-and-directors-who-cooperated-at-least-three-times/?envType=study-plan&id=sql-i)
:LOGBOOK:
CLOCK: [2023-04-10 Mon 20:22:06]
:END:
- LATER [1587. Bank Account Summary II](https://leetcode.com/problems/bank-account-summary-ii/?envType=study-plan&id=sql-i)
- LATER [1084. Sales Analysis III](https://leetcode.com/problems/sales-analysis-iii/?envType=study-plan&id=sql-i)
-
- DONE 复习数据库
- DONE 做 leetcode 上面的题 ((6433ff05-fb6e-4a21-9bef-37b8fdcff801))
:LOGBOOK:
CLOCK: [2023-04-11 Tue 09:27:22]--[2023-04-11 Tue 10:34:54] => 01:07:32
:END:
- DONE 做 quiz
-
- [[Open Music Theory]] [[Book notes]]
collapsed:: true
- Metadata:
- Time: 22:35
- Page: https://viva.pressbooks.pub/openmusictheory/chapter/introduction-to-western-musical-notation/
- Content:
- Answers for the Assignment:
- 1:
- #+BEGIN_VERSE
Rhythm and melody are the most important ones, the dynamics (forte, crescendo) is less important
#+END_VERSE
- 2:
- I would write down:
- Rhythm(including speed), and melody
- The instrument part (whether played on a piano or a guitar)
- Dynamics
- I would leave out:
- I might consider not putting a guitar tab if the staff is meant to played by multiple instruments
- 3:
- Yes, if I'm playing drums, I don't need melody.
- Or if the section of music is meant for improvisation, the rhythm and melody can be left out intentionally.
- 4:
- I will make a standard and explain what the notation means, and how my notation system is constructed.
-
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- 学数据库
- 看二手时间
- 跟朱思悦聊书
- 学习音乐
- What I've thought #thoughts
collapsed:: true
- 感觉我不应该这么强势，前几天否定了很多她喜欢的东西，不能让她因此就失去自尊，这样是一种不健康的关系。(虽然我感觉她还没有明显地表现出来）
- Mood
collapsed:: true
- 白天忘了，可能比较抑郁，因为上文所言，反正我现在感觉挺开心的 22:46

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- DONE 写区块链论文
SCHEDULED: <2023-04-13 Thu>
:LOGBOOK:
CLOCK: [2023-04-11 Tue 17:09:45]--[2023-04-11 Tue 17:37:59] => 00:28:14
:END:
- DONE 加思维导图
SCHEDULED: <2023-04-14 Fri>
- DONE 加引用
- DONE 写商谈通稿
:LOGBOOK:
CLOCK: [2023-04-11 Tue 21:18:32]--[2023-04-11 Tue 22:39:51] => 01:21:19
:END:
-
- DONE 再复习一次数据库
- DONE 看 ppt，找专业名词
- DONE 做点 leetcode 上面的题
-
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- 看书
- 学音乐
- 学校的东西
- What I've thought #thoughts
- 俄国文学看多了影响心理健康，[[二手时间]]里面各种惨案实在有点抑郁，要加强对极权的警惕
- Mood

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- Todo
- 乐理
- LATER 学 treble clef
- DONE 拿洗好的鞋
SCHEDULED: <2023-04-13 Thu>
- DONE 学数据库 lab
SCHEDULED: <2023-04-14 Fri>
:LOGBOOK:
CLOCK: [2023-04-13 Thu 09:15:57]--[2023-04-13 Thu 10:50:35] => 01:34:38
CLOCK: [2023-04-14 Fri 08:48:09]--[2023-04-14 Fri 08:48:14] => 00:00:05
CLOCK: [2023-04-14 Fri 08:50:52]--[2023-04-15 Sat 14:10:38] => 29:19:46
:END:

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- Todo
- DONE 修 neovim
- DONE 优化 neovim 启动速度 (失败)
- DONE 写概率论作业
:LOGBOOK:
CLOCK: [2023-04-13 Thu 13:57:23]--[2023-04-13 Thu 21:25:00] => 07:27:37
:END:
- DONE 28, 30,
:LOGBOOK:
CLOCK: [2023-04-13 Thu 14:12:56]--[2023-04-13 Thu 20:13:47] => 06:00:51
:END:
- DONE 33, 34,
- DONE 35, 39,
:LOGBOOK:
CLOCK: [2023-04-13 Thu 20:20:29]--[2023-04-13 Thu 20:41:11] => 00:20:42
:END:
- DONE 43, 51,
- DONE 52, 56
- DONE 准备军训的防晒
- LATER 存概率论的作业答案

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- LATER 把东西挂上咸鱼
:LOGBOOK:
CLOCK: [2023-04-15 Sat 19:01:21]--[2023-04-15 Sat 19:01:22] => 00:00:01
:END:
- LATER 二手书
- LATER 锅
- DONE 帆布包
- DONE 随身听
- LATER 想办法学英语口语
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- 军训
- 看书
- 家务
- What I've thought #thoughts
- 朱思悦一个周看完三本书比较 impressive，我觉得我也该挑战一下自己
- DONE 准备买阿兰德波顿和弗洛姆爱的艺术，这周没钱了
SCHEDULED: <2023-05-01 Mon>
- Mood

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- Leetcode - Non-overlapping Intervals
id:: 643d27e0-7330-4510-a2f7-8e2cecc08e7b
- Times:
- Time when completed: 10:33
- Time taken to complete: 30mins incl. learning
- DONE Revisions:
SCHEDULED: <2023-04-23 Sun>
:LOGBOOK:
CLOCK: [2023-04-16 Sun 10:33:49]--[2023-04-16 Sun 10:33:50] => 00:00:01
CLOCK: [2023-04-23 Sun 11:00:32]--[2023-04-23 Sun 11:09:38] => 00:09:06
:END:
- Tags:
- Algorithms: #greedy
- Data structures: #array #interval
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/non-overlapping-intervals/description/)
- [Previous solution](logseq://graph/logseq_notes?page=Leetcode%20Non-Overlapping-Intervals)
- Problem:
- Given an array of intervals `intervals` where `intervals[i] = [starti, endi]`, return _the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping_.
- Examples:
- ```
Example 1:
Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
Example 2:
Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
Example 3:
Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
```
- Constraints:
- `1 <= intervals.length <= 105`
- `intervals[i].length == 2`
- `-5 * 104 <= starti < endi <= 5 * 104`
- Thoughts:
- Intuition:
- This is a classic [[Interval Scheduling]] problem implemented using greedy algorithm.
- Approach:
- Sort the array by the end of interval
- Use greedy algorithm, find the interval with earliest finishing time
- Remove a interval if the starting time is before the finishing time
- otherwise, add the count
- return `length - count` for the intervals to be deleted
- Solution:
- Code
- ```java
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> a[1] - b[1]);
// Interval scheduling problem
int count = 1, minEnd = intervals[0][1];
// The first can be scheduled
for (int i = 1; i < intervals.length; i++) {
if (minEnd <= intervals[i][0]) {
// can be scheduled;
count++;
minEnd = intervals[i][1];
}
}
return intervals.length - count;
}
}
```

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- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- What I've thought #thoughts
- 今天一定要学乐理，这几天都没有学
- Mood
-
- [[Amused to Death]] [[Book notes]]
collapsed:: true
- Metadata:
- Time: 11:02
- Content:
- Teaching as an amusing activity 一章
- 经过我这不专业的观察，现在我们理工学校的文科政治老师都有一个趋势，就是要把教学娱乐化碎片化单一化，背后的目的是统一的，是为了更好地推进政府推崇的流氓思维，同时也阻抑你对政治局势的深入分析从而更方便听信政府的口径。
- 政治课让我最憋屈的一点就是上述的趋势，讲课这一活动从柏拉图/孔子式的多方交流思辨，变成了单方面的灌输。老师讲课中经常举例、归因，这本是文明人论述观点的常用手法，但和学术讨论的方式不同：学术讨论要求你点明从哪里引用，归因会给出大篇章节解释为什么这样归因，然而政治课不会给出任何引用来源和逻辑解释，老师信手拈来的观点和归因往往要靠你自己考证和推导，并且像老师这样高度娱乐化碎片化的教学设计导致的信息过载和中国传统不推崇反驳权威的社会风气，你没有机会反驳，你的反驳也不会被认可。这就导致老师上课随便扯美帝阴谋论和西方乱象的谣言，像我这样的人没空没能力没办法反驳的憋屈境况。
- 现在的感觉就是党为了推进这些思想，也不在意什么课程设计和教学的职业操守了，各种美帝阴谋论信手拈来，各种马克思习近平观点直接当成数学上的定理拿来就用，反正我把你这个设成必修课，你这牛马大学生想要毕业考研出国就必须要受我拿捏。
- 在 Reach out and Elect Someone 一章中
- 尼尔波茨曼说西方的世界正在走向赫胥黎所说的自愿独裁，中国和苏联这样的国家更像 1984 中的强力式独裁。而我感觉现在的中国更像是两者兼备，一方面用武力镇压游行虐待示威者来杀鸡儆猴，一方面靠操纵舆论，把舆论导到无关紧要的娱乐话题，在政治话题中搅混水，一方面让群众失去辨别真伪的逻辑思辨能力，一方面让民众沉湎于短平快娱乐失去思考政治的耐心，到最后就可以实现[[美丽新世界]]那样的自愿放弃权利，自愿被压榨做顺民
-
- todo
- DONE 修 logseq
- DONE darktable 颜色管理
- DONE 做作业
:LOGBOOK:
CLOCK: [2023-04-17 Mon 19:05:29]--[2023-04-17 Mon 22:57:37] => 03:52:08
:END:
- DONE 1,2
- DONE 4
:LOGBOOK:
CLOCK: [2023-04-17 Mon 20:13:00]--[2023-04-17 Mon 20:16:31] => 00:03:31
:END:
- DONE 6
- DONE 7,9
- DONE 上课问题(c)
- DONE 剪手指甲
- DONE 剪脚趾甲
- DONE leetcode
- DONE 领话费
SCHEDULED: <2023-04-18 Tue>
- LATER 写 leetcode 的笔记
- DONE 开加湿器

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- Leetcode - Set Matrix Zeroes
collapsed:: true
- Times:
- Time when completed: 08:51
- Time taken to complete: about 30 mins with answer
- Revisions:
- Tags:
- Algorithms: #array_in_place_operation
- Data structures: #vector_2d #array
- Difficulty: #difficulty_medium
- Platforms: #leetcode
- Links:
- link to the problem
- Problem:
- to-be-filled
- Examples:
- ```
```
- Constraints:
- to-be-filled
- Thoughts:
- Intuition:
- Approach:
- Solution:
- Code
-
- Todo
- LATER https://github.com/iamcco/markdown-preview.nvim
- DONE 带有辅助线的五线谱识别作业
-
- 书单
- DONE todo: 把新书加上去
-
- DONE 学数电 lab
SCHEDULED: <2023-04-19 Wed>

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- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- 白天学 lab，然后看《追忆似水年华》
- 晚上听音乐
- What I've thought #thoughts
- 很喜欢贝斯，想弹
- 花了太多钱，不该花这么多钱吃饭
- Mood
- 感觉还可以

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- DONE 刷智慧树网课 （等 dns 缓存刷新）
SCHEDULED: <2023-04-22 Sat>
:LOGBOOK:
CLOCK: [2023-04-20 Thu 08:44:46]
:END:
- DONE 修 sunshine
- DONE 键盘 caps-ctrl 互换
- LATER 研究 latex
-
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
collapsed:: true
- 学习
collapsed:: true
- 因为想政治的原因不是很专注，两个小时才把作业写完。
- 看书
- 玩游戏
- 看电影：瞬息全宇宙
- What I've thought #thoughts
collapsed:: true
- 学概率论之前看了下活吧，了解到一个 fact 说高考无法提供什么社会流动性，只是一个幻觉，就像科举一样，再次应证我的猜想：阶级和金钱才能决定你的社会生活是怎样。
- 之前一直认为恋爱要有信念，但我就属于那种信念不坚定的人，光环散去，觉得她不完全是我理想的那种类型，现在就随缘了，看社交 flow 行事，单不单身都无所谓了，总之要有自我的成长。
- Mood
collapsed:: true
- 挺好的
-
- [[拥抱逝水年华]] [[Book notes]]
collapsed:: true
- Metadata:
- Time: 16:16
- Page: 第四章
- Content:
- 看完第四章，总结而言就是要诚实面对自己的内心。面对自己的无知，面对自己的嫉妒，面对自己的愤懑，找到问题的根源，由此出发悟出更深入的道理，抑或是将自己尚且弱势的一面坦诚告人，继续寻找解决／提升的办法。（德波顿总结得和我个人的哲学很接近）
- [[拥抱逝水年华]] [[Book notes]]
collapsed:: true
- Metadata:
- Time: 16:47
- Page: 第五章 传情达意
- Content:
- 看这一章也印证了我的一个喜好就是不喜欢用网络用语。
- 那种玩烂梗的网络用语不用多讲，值得讨论的是那种形容得精妙恰当，不失风趣的修辞。书中提到早在 17 世纪法国中产盛行使用陈词滥调的修辞手法，例如形容“夕阳像火烧一样”，“月亮娇羞地散发光芒”（当然，这些对于我这样一个现代中国人来说还算不上陈词滥调），德波顿、我、当然还有普鲁斯特对此一致提出反对意见，其中的原因德波顿阐释了几种，但归纳下来就是一个：伤害真实表达。
- 往细了讲，用陈词滥调的第一个危害是影响社会的公共语言。说夕阳像火烧一样固然没错而且还很浪漫，然而所有人都这样说就让表达显得单一化。生活中的夕阳千变万化，岂是能用一个句式能搞定的？如此这般，扶植了这一个表达，压抑了其他表达，让写作枯萎失去生机。
- 从这一点延伸，可以得出第二点：不方便作者的自我表露。德波顿说普鲁斯特喜欢表达个人化且好的作家，我的理解就是写作是一件私人的事情。按照“对自己诚实”的哲学信条，陈词滥调对表现出自己的情感与喜好没有任何助益。相反，个人化的表达让你更好地流露自己的感情，例如称自己的亲密伴侣为宝贝，父母给孩子起关于小动物的爱称，都可以用来支持我的观点。
- 再延伸出来，我认为陈词滥调和个人表达的占比可以用作检验社会文化自由度的一把标尺。我相信只有百花齐放了，真正美好经典的作品才能够沉淀下来，换而言之，一个持续前进的整体社会环境离不开公民个人的努力和创造，而不该是只有一个权威指定好和坏。
- [[拥抱逝水年华]] [[Book notes]]
collapsed:: true
- Metadata:
- Time: 21:12
- Page: 第六章：交友之道
- Content:
- 最重要的一句话是友情和诚实不可得兼，意思是两者不可能同时完全达到。
- 德波顿提到，为了维护友情，我们会牺牲一部分自我，也会眼高手低地赞扬对方，给对方一个面子。而社交中快节奏、期望直接回应的机制也不适合深度思考和讨论。
- 而为了补足在交友中失去的自我和诚实，我们使用独处，写作等艺术方式探索内心。读和写的时候不需要我们考虑到对方的想法，可以反复斟酌，多次修改。这样有了良好的环境、充足的时间我们才能充分整理自己的想法。
- 所以说要做到两点：
- 认识到水至清则无鱼，要维护好人际关系，不可避免会遇到“虚伪”的一面
- 平衡友情中虚伪和诚实的关系。虚伪不等同于虚假，少一部分诚实也不意味完全不坦诚，要判断什么时候哪个该占上风，才能在社交中如鱼得水，左右逢源。

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- 中国地下乐队
collapsed:: true
- 退格 backspace (psychedelic rock
- 兵马司（发行）
- Re tros (daft funk
- Dream can
- Leetcode - Rotate Image
id:: 64422d1f-5db3-435d-ab4e-2b9fe5fd47cd
collapsed:: true
- Times:
- Time when completed: 14:28
- Time taken to complete: 15 mins ( with solution in mind )
- DONE Revisions:
SCHEDULED: <2023-05-26 Fri>
:LOGBOOK:
* State "DONE" from "LATER" [2023-04-28 Fri 16:45]
* State "DONE" from "LATER" [2023-05-23 Tue 20:26]
:END:
-
- Tags:
- Algorithms: #math
- Data structures: #array
- Difficulty: #difficulty_easy
- Platforms: #leetcode
- Links:
- [link to the problem](https://leetcode.com/problems/rotate-image/description/)
- Problem:
- You are given an `n x n` 2D `matrix` representing an image, rotate the image by **90** degrees (clockwise).
- You have to rotate the image [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm), which means you have to modify the input 2D matrix directly. **DO NOT** allocate another 2D matrix and do the rotation.
- Examples:
- https://assets.leetcode.com/uploads/2020/08/28/mat1.jpg
- https://assets.leetcode.com/uploads/2020/08/28/mat2.jpg
- ```
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]
Example 2:
Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
```
- Constraints:
- `n == matrix.length == matrix[i].length`
- `1 <= n <= 20`
- `-1000 <= matrix[i][j] <= 1000`
- Thoughts:
- Intuition:
- The solution is tricky, which requires you to use math.
- In order to do this, first, transpose, then reflect the matrix.
- Solution:
- java
- ```java
class Solution {
private void swap(int[][] matrix, int a, int b, int x, int y) {
int temp = matrix[a][b];
matrix[a][b] = matrix[x][y];
matrix[x][y] = temp;
}
public void rotate(int[][] matrix) {
// transpose and reflect
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < i; j++) {
swap(matrix, i, j, j, i);
}
}
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix.length / 2; j++) {
swap(matrix, i, j, i, matrix.length - j - 1);
}
}
}
}
```
- DONE 购物清单 [[Shopping lists]]
- DONE 香熏/香氛
- DONE 耳塞
- DONE 查英语成绩
SCHEDULED: <2023-04-25 Tue>
- DONE 报名
SCHEDULED: <2023-05-08 Mon>
- DONE 写乐理练习的作业

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- DONE 卖海底捞的锅
- [[拥抱逝水年华]] [[Book notes]]
collapsed:: true
- Metadata:
- Time: 21:05
- 章节: 九弃书不观
- Content:
- 弃书不观一章中，德波顿回应了我持久以来的一个困惑：我会成为书呆子吗？我该如何避免成为这样的人？简而言之：对书本的正确态度应是既能领略其妙处，又能察觉其限制。
- 讲到书的妙处，我联想到 [[华氏451]]中的一个观点，就是其中一个角色说读书应该做到三件事：
- 确保你有时间和精力读一本书
- 确保你有足够的时间想这一本书
- 确保你在读完这本书之后有一些行动
- 此章节就部分呼应了华氏中的第三点：书不是你精神生活的全部，相反，它是用来补足你的精神生活的。爱情小说中男女主的感情再撕心裂肺一波三折，都无法代替我自己的感情经历，而相反，我用局外人的视角观察小说中的人物，能更好理解我自己的经历，从而更好地丰富我自己的生活。更加形而上的文章也应如此，我分析作者的观点将大师的功力内化，这样才能拨云见日看清之前所困扰的谜题。
- 同时，书和其他艺术形式还有一个好处就是让我们重新看到生活中被忽略的美感。这一观点在“心胸豁然”一章也有提及。人们被“大师”被消费主义有意无意地迷惑，将别人捕捉到和塑造出来的美误认为是生活中美的全部，而生活的美不局限在什么艺术展音乐剧中，早起时听到的鸟鸣，雨后细碎的积雨云和在网上听到的演出现场一样很美，一样值得我们欣赏。
- 而德波顿同时也提出书的限制，和什么时候该“弃书不观”。他提到尽信书本会让人们出现许多症候：
- 视作家为神明，或者更概括地：艺术上的偶像崇拜
- 很明显地，这样会让人们的视野局限在前人有限的创作中，而生活的美是无处不在的。一个作家好，但并不能替代世界上其他的作家，一类艺术作品中的内容能够激起你的感情，不意味着它是绝对的至味，也不意味着有其他美好的食物。而普鲁斯特也有言：“一幅画的美并不取决于它画的是什么东西”。在我看来，艺术的表达方式固然重要，但其内核依然是表现的内容：体现了人类怎样的普世问题？艺术家通过这幅画想要告诉你或给你展示什么？而画的内容，只不过是上述一切都载体。
- 看罢好书不能停笔
- 这个问题的来源和处方也很明显，因为我认为每个人的个人经验都是不可替代的，普鲁斯特写人的年华易逝，并不会影响白先勇写台湾人国族认同以及国民党人对于江川易主的感慨。
- [[拥抱逝水年华]] [[Book notes]]
collapsed:: true
- Metadata:
- Time: 19:44
- Page: 第八章：享受爱情
- Content:
- 德波顿在此章分析了爱情的两个方面：性交和吸引力
- 在吸引力方面，德波顿分析指出相爱的人很容易对对方感到厌倦，吸引力会下降。原因很明显，人是习惯的动物。德波顿为了解释相爱之人在相处久后自认为双方十分了解但实际上不是的现象（参见[[亲密关系]]）引用了诺亚的例子：诺亚活了500年，看尽人间草木，但当他被关在方舟同花鸟鱼兽作伴40天，出来发现世界面目全非之后，对世间万物的怜惜之情油然而生。亲密关系也有类似的论述，在恋人正要离你而去时，你才能回想起跟她一起幸福生活的点点滴滴（除非你在此时已经变了心）。
- 同时，德波顿也列举了维护吸引力的一些方法：
collapsed:: true
- 延迟享受：德波顿举了富人和穷人的例子：富人的富算是一种诅咒，因为他想要什么立马就会得到，而得到之后很快就会厌倦，而穷人对一件商品朝思暮想，得到后更加满足，因为自己为此付出了很多努力。 [[Apr 23rd, 2023]]更新：这不代表要故意冷落他人，根据[[亲密关系]]引用到的研究，故意冷落潜在对象对忠诚的影响是**负相关**
- 利用第三者：根据恋爱的三角关系理论： 欲他人之所欲，这样可以有效拉近双方的距离。我个人认为这种方式有可能玩过火，其他维护吸引力的方法可参见[[亲密关系]]
- 在性交方面，普鲁斯特解答了一个中国人很爱问的问题：是否支持婚前性行为。在他的观念里，性行为是为了爱情而服务的。跟上述关于吸引力的论述一样，性行为同样是和“延迟享受”紧密相连的，他指出男人嫖娼不可取，原因有二：
collapsed:: true
- 一方面是因为和良家妇女不同，妓女为了营业不可能拒绝，你从和妓女性交中得到的快感并不能持续。
- 第二方面也是我认为更重要的方面：嫖客以为自己从性交中得到了爱情，但很明显这只是对于爱情的幻觉。而像这样从射精到感觉到“爱”的条件反射反而会抑制一个人对真正爱情的追求。他会被这个幻象迷惑，认为自己不满足是因为性能力不够，对方不配合，但实际上不满足的结果更指向问题本身：想要从嫖娼中得到爱情。
- 尽管我不嫖娼，德波顿这里关于嫖娼的叙述依然值得我深思，它反映的是资本主义社会中现代人的“代餐”文化：用高潮代替爱情，用披着教育性外壳的娱乐视频代替深度的系统性的学习，用肤浅的消费主义代替幸福。应该常常质问自己：这（自慰、看油管视频、暴食、冲动消费）是不是我真正想要的？
-
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- 军训
- 花一晚上读书和写读书笔记
- What I've thought #thoughts
- 写读书笔记是真的能促进专注力，一坐下去一晚上就没了
- DONE 今天主要是比较累，而且没想到写读书笔记能这么认真，每日任务没做什么，所以明天一早就要去把每日任务做了。
SCHEDULED: <2023-04-23 Sun>
- Mood
- 挺开心的反正
-
- 书籍元数据:
template:: 书籍元数据
- 作者：
- 阅读时间：
- 评分：? / 5 [[Book reviews]]
- 评价内容

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- DONE 联系那个什么比赛的什么人（不想联系他了）
- DONE 学概率论
- LATER #抖音赚钱计划 #projects
:LOGBOOK:
CLOCK: [2023-05-11 Thu 10:22:24]--[2023-05-11 Thu 10:22:25] => 00:00:01
CLOCK: [2023-05-11 Thu 10:22:26]--[2023-05-11 Thu 10:22:27] => 00:00:01
:END:
- DONE 一台二手机（我的索尼手机）
- DONE 手机号（可以用我自己的手机）
- 视频素材
- 游戏
- DONE lol
- DONE (不要了)王者荣耀
- DONE MC跑酷
- 正经
- LATER 电影
:LOGBOOK:
CLOCK: [2023-05-08 Mon 19:38:12]--[2023-05-08 Mon 19:38:13] => 00:00:01
:END:
- LATER TED演讲
- LATER 采访
- LATER 音乐演出
- 音频
- LATER 音乐演出
- LATER 油管流行音乐
- 技术
- LATER scripting kdenlive
- 账号
- DONE 抖音账号
- LATER youtube shorts

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- 产开论文
collapsed:: true
- 内容
- DONE (辛宇) 机会识别 <- 发展现状，分析和痛点
background-color:: yellow
- DONE (胡恒瑞) 客户需求识别
background-color:: red
- 有一个五步流程：缝合一个老人和残疾人上去
- DONE (李文状) 使命宣言
background-color:: pink
- 在产品开发中，使命宣言是指明公司、团队或产品的愿景和目标的声明。它描述了公司或产品的核心目的和价值主张，并概括了公司或产品为客户或社会带来的益处。使命宣言的编制过程需要深入了解公司、团队或产品的定位、竞争环境和目标客户群体，以确保它准确地反映了公司或产品的核心特点和差异化优势，并且可以激励团队向着共同的目标努力。在整个产品开发过程中，使命宣言可以作为一个指导方针，指导团队在产品设计、开发和营销等方面做出决策，并确保产品始终与公司的核心价值主张保持一致。
- DONE 产品概念 (辛宇) <- 四、区块链在物流行业中的应用
background-color:: yellow
- DONE 知识产权战略 (胡恒瑞)
background-color:: red
- DONE 结论（李文状）
background-color:: pink
- 结构 (黄源源、张富洋)
- DONE 原创性 (查重)
- DONE 整体质量（格式）
- DONE 2500词以内
- DONE 其他要求
- ![image.png](../assets/image_1682336346047_0.png)
- DONE 团体流程表
-
- DONE 智慧树的见面课作业
SCHEDULED: <2023-04-28 Fri>
-
- 书单 #books [[Book lists]] #excluded_in_query
- 哲学
- 亲密关系、女性主义
- LATER [[简明哲学导论]]
- LATER [[性差异的伦理学]]
- LATER [[爱为什么痛]]
- 政治学
- LATER [[共同的底线]]
- 逻辑学
- LATER [[好好讲道理]]
- 文学
- LATER [[怨女]]
- DONE [[倾城之恋]]
- LATER [[让我独自一人]]
-
- 北京日料餐厅 #food
- 居酒屋
- LATER 慢走
- 鳗鱼饭
- LATER 傲鳗
- 烤猪排
- LATER 虎铁
- DONE 圣博殿
- 托付料理
- LATER 雪崴
- LATER 凤寿司

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- Leetcode - Word Search
collapsed:: true
- Times:
- Time when completed: 10:47
- Time taken to complete: 30mins
- DONE Revisions:
SCHEDULED: <2023-05-30 Tue>
:LOGBOOK:
* State "DONE" from "LATER" [2023-05-11 Thu 20:29]
* State "DONE" from "LATER" [2023-05-23 Tue 20:26]
:END:
- [[May 11th, 2023]] 11:43
- Tags:
- Algorithms: #DFS #DFS_preorder #backtrack
- Data structures: #vector_2d
- Difficulty: #difficulty_medium
- Platforms: #leetcode
- Links:
- [Link to the problem](https://leetcode.com/problems/word-search/description/)
- Problem:
- Given an `m x n` grid of characters `board` and a string `word`, return `true` *if* `word` *exists in the grid*.
- The word can be constructed from letters of sequentially adjacent
cells, where adjacent cells are horizontally or vertically neighboring.
The same letter cell may not be used more than once.
- **Follow up:** Could you use search pruning to make your solution faster with a larger `board`?
- Examples:
- https://assets.leetcode.com/uploads/2020/11/04/word2.jpg
- https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg
- https://assets.leetcode.com/uploads/2020/10/15/word3.jpg
- ```
Example 1:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true
Example 2:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true
Example 3:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false
```
- Constraints:
- `m == board.length`
- `n = board[i].length`
- `1 <= m, n <= 6`
- `1 <= word.length <= 15`
- `board` and `word` consists of only lowercase and uppercase English letters.
- Thoughts:
- Intuition:
- The solution is simple, first loop over the board to find a starting point, then start dfs there.
- The only caveat is remember to add a simple backtracking to prevent the case when a route is invalid, but still used the visited[][] array.
- Approach:
- Loop over the board
- DFS from that point
- Solution:
- Code
- ``` java
class Solution {
boolean[][] visited;
boolean dfs(char[][] board, String word, int x, int y, int start) {
if (start == word.length()) {
return true;
}
if (x < 0 || x >= board.length || y < 0 || y >= board[0].length) {
return false;
}
if (visited[x][y]) {
return false;
}
if (board[x][y] == word.charAt(start)) {
visited[x][y] = true;
boolean found = dfs(board, word, x + 1, y, start + 1) ||
dfs(board, word, x - 1, y, start + 1) ||
dfs(board, word, x, y + 1, start + 1) ||
dfs(board, word, x, y - 1, start + 1);
if (!found) { // the backtracking
visited[x][y] = false;
return false;
} else {
return true;
}
} else {
return false;
}
}
public boolean exist(char[][] board, String word) {
visited = new boolean[board.length][board[0].length];
// first, loop over the problems, then use dfs.
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (word.charAt(0) == board[i][j]) {
visited = new boolean[board.length][board[0].length];
if (dfs(board, word, i, j, 0)) {
return true;
}
}
}
}
return false;
}
}
```

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- DONE 看见面课
- DONE 催杨哥改稿

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- LATER 看见面课

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- 五一成都旅行计划
- [[May 1st, 2023]]
- 早上吃早餐，弹贝斯看书
- 中午去万象城润街逛吃
- 下午去东郊记忆逛
- 晚上去东郊记忆livehouse
- [[May 2nd, 2023]]
- 下午去icd看电影 (16:55 - 18:28)
- 晚上在icd吃饭
- [[May 3rd, 2023]]
- 白天骑绿道踏青（笑死这是夏天
-

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- #projects [[健身]]
- 增重增肌
- Workouts:
- Starting: Three days / week
- https://steelsupplements.com/blogs/steel-blog/the-big-3-workout-routine-everything-you-need-to-know
- 哑铃卧推
- https://barbend.com/dumbbell-bench-press/
- A little advanded: Three days / week
- File: ![mnsfullbody.pdf](../assets/mnsfullbody_1683295663220_0.pdf)

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- LATER 作业
- DONE 看概率论网课
SCHEDULED: <2023-05-08 Mon>
:LOGBOOK:
CLOCK: [2023-05-06 Sat 14:47:29]--[2023-05-06 Sat 20:11:21] => 05:23:52
:END:
- LATER 写概率论ch3作业
- DONE 11, 12, 14, 15
- LATER 18, 20
- LATER 21, 24, 25
- DONE 报名雅思
SCHEDULED: <2023-05-08 Mon>
- DONE 整理书单
SCHEDULED: <2023-05-07 Sun>
- DONE 学数据库、java
SCHEDULED: <2023-05-08 Mon>
- DONE java lab
:LOGBOOK:
CLOCK: [2023-05-07 Sun 11:04:14]--[2023-05-07 Sun 11:49:33] => 00:45:19
:END:
- DONE 数据库第三周
- DONE 数据库lab
- DONE Logseq：学datalog
SCHEDULED: <2023-05-09 Tue>
- DONE 修anki雅思单词的发音：改天换新的
- DONE 尝试健身
- DONE 找bench push up的替代方式
:LOGBOOK:
CLOCK: [2023-05-06 Sat 20:01:17]--[2023-05-06 Sat 20:01:20] => 00:00:03
:END:
- 常规push up

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- DONE 学 #健身
- DONE deadlift硬拉
- DONE squat深蹲
- LATER push up 俯卧撑
- DONE Gnome 拓展 (不行)
SCHEDULED: <2023-05-11 Thu>
:LOGBOOK:
CLOCK: [2023-05-11 Thu 15:20:16]--[2023-05-11 Thu 15:20:17] => 00:00:01
:END:
- DONE gnome per keyboard config

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- DONE 在电脑上下anki单词包
- LATER neodb
- LATER 书评
- LATER 影评
- DONE 实验课
- DONE 配置环境
- LATER 科研
- LATER 看一下形势
- #projects 赛博形象管理 #latex
- LATER 简历
- LATER 含个人信息版本
- DONE 中文版
- LATER 英文版
- LATER 不含个人信息版本
- LATER 能力
- LATER 会latex
- LATER 会剪视频
- LATER 个人网站
- LATER 把以前的移到其他地方
- LATER 更新网站，用hugo之类的cms

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- <% today %> #健身 #健身记录
template:: workouts
collapsed:: true
- Deadlift
- 7.5 kg
- 5 reps
- 3 sets
- Squat
- 7.5 kg
- 5 reps
- 3 sets
- Dumbbell Bench Pushup
- 5 kg
- 10-15 reps
- 3 sets
-
- [[May 10th, 2023]] #健身 #健身记录
collapsed:: true
- Deadlift
- 7.5 kg
- 5 reps
- 3 sets
- Squat
- 7.5 kg
- 5 reps
- 3 sets
- Dumbbell Bench Pushup
- 5 kg
- 10-15 reps
- 3 sets
-

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- DONE [#A] 复习java
- DONE 看课件
- DONE 复习lab
- DONE tutorial (看到q6)
:LOGBOOK:
CLOCK: [2023-05-11 Thu 21:03:15]--[2023-05-11 Thu 21:33:21] => 00:30:06
:END:
- LATER [#C] 复习数据库
- LATER 看课件
- LATER 做作业
- LATER [#C] #projects #抖音赚钱计划
- LATER 电影剪辑
- LATER 片源
- LATER Youtube
- LATER IMDB
- LATER [#C] 实验报告
- LATER 学latex

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- [[May 12th, 2023]] #健身 #健身记录
collapsed:: true
- Deadlift
- 7.5 kg
- 5 reps
- 5 sets
- Squat
- 7.5 kg
- 5 reps
- 5 sets
- Dumbbell Bench Pushup
- 5 kg
- 10 reps
- 3 sets
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- 玩王国之泪
- 健身
- 看了点java lab作业
- 看一亩三分地
- What I've thought #thoughts
- 昨天没睡好，今天就算健了身也有点无精打采，玩了一天游戏，也没学什么
- 看地里面感觉有点焦虑，目前对于美利坚贴金gpa还不够，虽然招工可以了
- 刷完智慧树，少一件事可以焦虑
- Mood

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- #projects [[New computer 2023]]
- OS
- https://liguros.gitlab.io/
- Daily reflection [[Daily reflections]]
- What I've done
collapsed:: true
- 玩王国之泪
- 和鲁、于吃晚饭，和鲁看电影《在路上》，然后和鲁一起回学校
- What I've thought #thoughts
collapsed:: true
- 她接受了我的未来规划，这一点我觉得很好，但她自己没有什么对未来的规划，有时间应该帮她看一下
- Mood
collapsed:: true
- 满足且快乐

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- [[May 15th, 2023]] #健身 #健身记录
- DONE Lateral raise
- DONE 2kg wired
- DONE 15 reps
- DONE 2.5kg dumbbell
- DONE 15reps
:LOGBOOK:
CLOCK: [2023-05-15 Mon 10:30:47]--[2023-05-15 Mon 10:30:48] => 00:00:01
:END:
- DONE Squat
- DONE 7.5 kg
- DONE 5 reps
:LOGBOOK:
CLOCK: [2023-05-15 Mon 10:05:52]--[2023-05-15 Mon 10:05:53] => 00:00:01
:END:
- DONE 2 sets
- DONE 10 kg
- DONE 5 reps
- DONE 3 sets
- DONE Dumbbell Bench Pushup
- DONE 7.5 kg
:LOGBOOK:
CLOCK: [2023-05-15 Mon 10:19:37]--[2023-05-15 Mon 10:19:38] => 00:00:01
CLOCK: [2023-05-15 Mon 10:19:39]--[2023-05-15 Mon 10:19:40] => 00:00:01
:END:
- DONE 15 reps
- DONE 3 sets
- DONE Barbell Row
:LOGBOOK:
CLOCK: [2023-05-15 Mon 10:12:58]--[2023-05-15 Mon 10:12:59] => 00:00:01
:END:
- DONE 5 kg
- DONE 5 sets
- DONE 5 reps
- DONE Pull-up
- DONE 40.8kg
- DONE 1 set
- DONE 8 reps
- DONE 49.9kg
- DONE 1 set
- DONE 8
- DONE Sit-up
- DONE 5 sets
:LOGBOOK:
CLOCK: [2023-05-15 Mon 10:29:08]--[2023-05-15 Mon 10:29:09] => 00:00:01
CLOCK: [2023-05-15 Mon 10:33:11]--[2023-05-15 Mon 10:33:12] => 00:00:01
CLOCK: [2023-05-15 Mon 10:33:36]--[2023-05-15 Mon 10:33:37] => 00:00:01
:END:
- DONE 6 reps
-
- Daily reflection [[Daily reflections]]
- What I've done
collapsed:: true
- 没睡好，睡了六个小时
- 学概率论，挺恶心的
- What I've thought #thoughts
collapsed:: true
- 鲁有点反复无常，挺烦的，现在先摆烂了
- Mood
- 累
-
-

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- #projects #健身
- 5x5 workout
- https://www.bodybuilding.com/content/5-best-bodybuilding-programs-to-pack-on-serious-muscle.html

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- LATER 学习任务
- DONE 概率论
- LATER 作业
- ch3 27 31 36 37 39 41 42 43
- ch4 12 13 15 23
- ch5 3 6 7
- DONE 数字电路

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- #projects #健身 #健身计划
- 5 x 5 workout
- Perform workout A twice, on Monday and Friday. Perform workout B once on Wednesday.
- **Week 1**
- Workout A #健身记录 #健身
template:: 5x5 workout week1 a
- DONE Barbell back squat — 5×5
- DONE Dumbbell bench press — 15x3
- DONE Barbell row — 5×5
- Workout B #健身记录 #健身
template:: 5x5 workout week 1 b
- DONE Barbell back squat — 5×5 (10kg x 3, 7,5kg x 2)
- DONE Dumbbell shoulder press — 3x10 (5kg)
- DONE Barbell deadlift — 1×5 (10kg)
:LOGBOOK:
CLOCK: [2023-05-19 Fri 11:08:53]--[2023-05-19 Fri 11:08:55] => 00:00:02
:END:
- **Week 2**
- Workout A
- Barbell back squat — 5×5
- Barbell overhead press — 5×5
- Barbell deadlift — 1×5
- Workout B
- Barbell back squat — 5×5
- Barbell bench press — 5×5
- Barbell row — 5×5
- Week 3 will repeat the structure of week 1, adding weight. Week 4 will follow week 2’s structure, adding weight.
-
- 把健身记录写在一个表格上
SCHEDULED: <2023-05-20 Sat>

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- LATER 概率论
:LOGBOOK:
CLOCK: [2023-05-19 Fri 13:26:19]
CLOCK: [2023-05-19 Fri 13:26:31]--[2023-05-19 Fri 13:26:45] => 00:00:14
:END:
- DONE 电路
:LOGBOOK:
CLOCK: [2023-05-19 Fri 13:26:31]--[2023-05-19 Fri 13:26:45] => 00:00:14
:END:
- DONE 宿舍文化节
:LOGBOOK:
CLOCK: [2023-05-19 Fri 13:26:29]
CLOCK: [2023-05-19 Fri 13:26:34]
CLOCK: [2023-05-19 Fri 13:26:38]--[2023-05-19 Fri 13:26:44] => 00:00:06
:END:
- DONE 形势与政策论文

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- Daily reflection [[Daily reflections]]
- What I've done
collapsed:: true
- 看书
- 学概率论
- 和鲁学习吃饭
- What I've thought #thoughts
collapsed:: true
- 今天说了很多话，跟鲁讲了很多社会现象和我的分析并获得认可，也听她说了自己的看法，很好
- Mood
collapsed:: true
- 好

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- Daily reflection [[Daily reflections]]
- What I've done
- 上午研究抓包，认证802.1x PEAP
- 下午玩王国之泪
- 晚上写毛概，背单词，写leetcode
- What I've thought #thoughts
-
- Mood
- 开心

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- DONE 写形势论文

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- DONE 形象管理
- DONE 剪指甲
- DONE 剃胡子
- DONE 研究香水

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- 今日任务 [[学习任务]]
id:: 646ef30b-9a68-4c08-87a2-e5d9ca0b5f81
- DONE 学java
- DONE 看课件，学知识点
- DONE 看project要求
- DONE 写project
:LOGBOOK:
CLOCK: [2023-05-25 Thu 18:34:56]--[2023-05-25 Thu 18:34:57] => 00:00:01
:END:
- DONE Manual
SCHEDULED: <2023-05-30 Tue>
- DONE 概率论
- DONE 写作业
- DONE 看课件
:LOGBOOK:
CLOCK: [2023-06-03 Sat 16:44:31]--[2023-06-03 Sat 16:47:00] => 00:02:29
:END:
- DONE 看网课
- DONE 数据库
- DONE 看要求
- DONE 做视频
- DONE 拍摄
SCHEDULED: <2023-05-26 Fri>
- DONE 剪辑
SCHEDULED: <2023-05-27 Sat>
- 该看点小说

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- Daily reflection [[Daily reflections]]
- What I've done
collapsed:: true
- 健身
- 王国之泪
- What I've thought #thoughts
collapsed:: true
- 鲁对我爱理不理，我说话时机不好，她应该是有backup，whatever，有又怎样，没有又怎样。Follow the flow
- 还是该多看书，要一直提升自己的魅力。
- Mood
collapsed:: true
- 还行

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- 概率论
- 5.8, 5.10
- 6.3, 6.4(a), 6.5
-
- #概率论
- 复合函数期望公式
- ![2023-05-27-19-49-01.jpeg](../assets/2023-05-27-19-49-01.jpeg)
- Daily reflection [[Daily reflections]]
- What I've done
- What I've thought #thoughts
- 跟她聊天不能获得什么即时反馈，跟她的关系一直如此也没有进展，不知道是不是她故意晾着我，反正这几天都没睡好，要是再这样下去，没得到什么情感价值反而为了这些焦虑不如干脆算了或者换人。我现在决定将计就计，下决心最近都不和她交往了，看看她反应。
- Mood
-
-

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- DONE 修手柄
- ((646ef30b-9a68-4c08-87a2-e5d9ca0b5f81))
- DONE 写latex
:LOGBOOK:
CLOCK: [2023-05-29 Mon 18:01:16]--[2023-05-29 Mon 18:01:17] => 00:00:01
:END:
- java大作业

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- [[学习任务]]
- DONE 预习lab
SCHEDULED: <2023-05-30 Tue>
- DONE 背马原
SCHEDULED: <2023-06-01 Thu>
-
- [[New computer 2023]]
- Userland software for asus laptops (gentoo repo)
- https://lab.retarded.farm/zappel/zGentoo
- don't enable global ~amd64, gnome 44 is not stable.

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- [[学习任务]]
- DONE 背产开
- DONE 下载课件
- LATER 写思维导图
:LOGBOOK:
CLOCK: [2023-06-01 Thu 17:00:59]--[2023-06-01 Thu 17:01:01] => 00:00:02
:END:
- Exercises

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- [[学习任务]]
- LATER 概率论 {{renderer :todomaster}}
id:: 647bf024-41f1-45e7-afaf-f49026e826d6
:LOGBOOK:
CLOCK: [2023-06-03 Sat 16:47:39]--[2023-06-03 Sat 16:47:40] => 00:00:01
:END:
- LATER 学积分
:LOGBOOK:
CLOCK: [2023-06-03 Sat 17:05:07]--[2023-06-03 Sat 17:05:08] => 00:00:01
CLOCK: [2023-06-03 Sat 21:07:06]--[2023-06-03 Sat 21:07:07] => 00:00:01
CLOCK: [2023-06-03 Sat 21:07:17]--[2023-06-03 Sat 21:07:18] => 00:00:01
:END:
- DONE 网课
:LOGBOOK:
CLOCK: [2023-06-03 Sat 21:07:14]--[2023-06-03 Sat 21:07:20] => 00:00:06
CLOCK: [2023-06-03 Sat 21:07:21]--[2023-06-03 Sat 21:07:22] => 00:00:01
:END:
- LATER 写题（作业，对答案）
- LATER 总复习, 考试6.19，杰克6.8
- LATER 看完网课和串讲
- LATER 写思维导图
- LATER 做题
- DONE 区块链：实验报告
- DONE java,下周小测 [qm website](https://qmplus.qmul.ac.uk/mod/page/view.php?id=2178228)
SCHEDULED: <2023-06-07 Wed>
- DONE 课件
- DONE lab
- DONE lab6
- DONE lab7
- DONE PDP quiz
SCHEDULED: <2023-06-15 Thu>
:LOGBOOK:
CLOCK: [2023-06-04 Sun 16:18:08]--[2023-06-04 Sun 16:27:03] => 00:08:55
:END:
-
- Daily reflection [[Daily reflections]]
collapsed:: true
- What I've done
- 早上看数据库
- 下午看概率论，看了点书
- 晚上荒野大镖客，看完概率论和数据库
- 发现洗的衣服忘拿了
- What I've thought #thoughts
- [[情感]]
- 这几天没有睡好，有点抑郁，一方面是因为生了病没睡好，但主要是因为我的异性关系没有处理好。现在就很能共情[[老残游记]]里面的恋爱场景，一旦相隔远了没有消息，在床上翻来覆去睡不着的场景：一会觉得对方是好得极好，一会又觉得对方好像一无是处。我跟她也hang out了很多次了也进行了不少自我表露，但是那天我们突然就不再联系。
- 过去和现状：现在回忆起来，当时我是提出餐厅约会受拒，心情不好也顺便推掉了study date（不过她也没有如此的打算，和室友一起），后来遇见她也是跟着室友一起，会跟我打招呼，但是我几次找她，她都借故推脱，再之后就没有了联系。直到最近她主动跟我找话，但是因为之前她拒绝的时候同时拒绝了study date的主要工具，我也没有办法继续找借口交往。到现在我们在群聊等公众场合互相避让，不回复互相的信息。
- 抑郁的原因：
- 我被拒绝
- 我感觉被操纵，她对我的情感表露不真诚。感觉她和我的交往有很多套路性的元素，就像主动拒绝study date，故意秒回，故作高冷。这些方法在统计学学上只会降低对异性的吸引力。
- 之前通过线上途径交往十分不理想，人的需求不能得到及时满足，因为她会故意晚回我的消息（这一点不完全有把握，是因为看见她会秒回别人的消息
- 一直挂着没有联系，这一方面是因为二和三，一方面也是因为话题被她切断，没有办法
- 怎么办
- 到现在已经没什么主动办法，因为聊天的线索被切断，其他活动型的date她大概率不会接受，因为快要期末了
- 唯一的办法是一直憋着不表示，不明确拒绝她也不主动接近她，先让她急起来，主动放弃这些套路，然后再跟随着flow来
- 如果一周没有进展的话，我就应该主动放弃，尽量回避不能再这样互相消耗了
- Mood

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- [[Personal Tasks]]
- [[New computer 2023]]
- DONE 导入书签
- DONE 测试扬声器
- [[Learning Logseq]]
- LATER Learn datalog
-
- Daily reflection [[Daily reflections]]
- What I've done
- Morning: learn to fix suspend, and other asus stuff
- Afternoon: Washed clothes, did java lab 6
- What I've thought #thoughts
- Random thing I learning while browsing the internet:
- Wear earplugs when in concert! [[music]]
- Mood

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- [[产品开发]]
- [product development process from shopify](https://www.shopify.com/blog/product-development-process)
- [Idea generation](https://www.shopify.com/blog/product-development-process#1)
- [Market Research](https://www.shopify.com/blog/product-development-process#2)
- [Planning](https://www.shopify.com/blog/product-development-process#3)
- [Prototyping](https://www.shopify.com/blog/product-development-process#4)
- [Sourcing](https://www.shopify.com/blog/product-development-process#5)
- [Costing](https://www.shopify.com/blog/product-development-process#6)
- [Commercialization](https://www.shopify.com/blog/product-development-process#7)
-
- Daily reflection [[Daily reflections]]
- What I've done
- 早上健身，传区块链虚拟机
- 下午玩喷3
- 晚上看书，看产开
- What I've thought #thoughts
- 今天因为没date上有点颓废，玩了半天游戏。明天不能这样了，明天去学习。
- Mood

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- [[New computer 2023]]
- LATER import ssh keys
- LATER home computer
- LATER cloud server
- LATER pad?

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- Today's tasks
- {{embed ((647bf024-41f1-45e7-afaf-f49026e826d6))}}
-
-
- Daily reflection [[Daily reflections]]
- What I've done
- 上午跟鲁交换学习资料
- 下午和晚上玩喷二
- What I've thought #thoughts
- Mood

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- [[学习任务]]
-
- Daily reflection [[Daily reflections]]
- What I've done
- Morning: Workout
- Afternoon: Review [[产品开发]] and [[Java]]
- Night: Review 数电 学交
- What I've thought #thoughts
- Mood
-
- [[产品开发]]
collapsed:: true
- ## Topic 6 - 10
- ## Topic 6 - Phase 1: Concept Development 概念开发
id:: 6482bafb-b96b-44c3-834b-8b4966c1dc0c
collapsed:: true
- What is Concept Development? 概念开发是什么
- The needs of the target market are identified
Alternative product concepts are generated and
evaluated 需求
One or more concepts are selected for further
development and testing 多种概念被生成，经过挑选后选择其中一个
A concept is a description of the form, function
and features of a product 概念是一个产品的描述，功能和特性
Evaluation and screening criteria are used to aid
in the selection 评估和测试可以辅助选择（screen = test）
Usually accompanied by a set of specifications, an
analysis of competitive products and an economic
justification for the project 伴随着来自个方面的要求，与竞争对手的比较和分析，还有经济方面的考虑（能不能赚钱）
- This can be concluded as 图解如下
- ![image.png](../assets/image_1686289806889_0.png)
- #### Identify Customer Needs 用户需求是什么
- to understand customers needs,
then to effectively communicate them to the
development team 和研发队伍沟通
The output of this step is:
Customer need statements organised in a
hierarchical list, with importance weightings
for many or all of the needs 把用户的需求整理成加权列表
- #### Establishing target specifications 确立规格要求
- 1. Prepare the list of metrics, i.e. the technical or
manufacturing features of the product based on
the customer needs 准备一些数据的列表
- 2. Collect competitive benchmarking information 竞争对手的产品性能
- 3. Set ideal and marginally acceptable target values 理想的和可接受的要求
- 4. Reflect on the results and the process 通过结果和进一步完善
- #### Concept generation 想一个概念
- Steps:
- external search 在外面搜索
- creative problem solving within the team, and 团队灵机一动
- systematic exploration of the various solution
fragments the team generates 系统地寻找解决方法
- The result of this activity is usually a set of 10-20
concepts – each is typically represented by a sketch
and a brief descriptive text 产生 10-20 个概念
- ![image.png](../assets/image_1686289726119_0.png)
- #### Concept selection 选一两个出来
- the activity in which various
product concepts are analysed and sequentially
eliminated to identify the most promising
concept(s)
- ![image.png](../assets/image_1686289907417_0.png){:height 564, :width 688}
- There are 5 stages to the screening and
evaluation process: 用以下方式选择好的概念
- 1. Initial screen - The key evaluation criteria given in the table on the
previous slides can be developed further using a
scoring model or weighted checklist 用之前的要求测试
- 2. Customer screen - An informal discussion with customers to
explain a concept 和客户沟通
- 3. Technical screen 在科技实现上咨询第一方和第三方的专家
- informal technical discussions with experts
- extensive analysis by a 3 rd party
- 4. Final screen 最终测试
- Involves the use of screening models and computer assessment programs
- 5. Business analysis 看下赚不赚钱
- preliminary marketing plans,
- technical plans,
- financial reviews and
- projected budgets
- #### Concept testing 测试
- There are 7 steps to this process
- 1. Define the purpose of the concept test 定义测试的目的
- 2. Choose a survey population 选问卷调查的目标对象
- 3. Choose a survey format 格式
- 4. Communicate the concept 和目标对象沟通这个概念
- 5. Measure customer response 客户怎么想
- 6. Interpret the results 分析结果
- 7. Reflect on the results and process 用结果改进
- #### Setting final specifications 最终规格
Target specifications are revisited after a concept has been selected and tested
- #### Project planning 计划
- The final activity of concept development
- Steps:
- creates a detailed development schedule 产品开发时间线
- devises a strategy to minimise development time and 定一个计划
- identifies the resources required to complete the project 需求是什么
- #### Economic analysis
- 拿出去卖亏不亏钱
- #### Benchmarking & modelling
- 和竞争对手比较，品质行不行
- ## 往年题考点
- ### Prototypes (2019)
- Physical prototype, (5 marks) (2019)
- Analytical prototype, (5 marks) (2019)
- Comprehensive prototype. (5 marks) (2019)
- ### Costs (2019)
- Component costs, (5 marks) (2019)
- Support costs, (5 marks) (2019)
- Indirect allocations as part of Overhead costs. (5 marks) (2019)
- ### New products and New product development (2019, 2018)
- #### Booz, Allen & Hamilton (1982) Classification Scheme (2019)
- Repositioning (4 marks) (2019)
- four (4) elements through which products can be differentiated (2019)
- the five alternatives that are open to the firm ( 2018)
- kinds of new products identified by Booz Allen and Hamilton ( 2018)
- four elements where values can be added to the basic or ‘core’ product to augment its appeal ( 2018)
- Product life cycle (2019)
- #### prototyping ( 2018)
- three phases of prototyping ( 2018)
- uses of prototypes ( 2018)
- ### system-level design (2019)
- product architecture (2019)
- key characteristics of modular and integral architecture. (2019)
- three (3) main types of modular architecture. (2019)
- ### IP ( 2018)
- the concept ( 2018)
- what sort of intellectual property copyright can protect and what form of protection it gives. ( 2018)
- benefits and limitations of using patents ( 2018)

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- Daily reflection [[Daily reflections]]
- What I've done
- Morning: review DB
- Afternoon: review DB and Mao
- Night: review Mao and DB
- What I've thought #thoughts
- Mood

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- Daily reflection [[Daily reflections]]
id:: 64856c09-6826-4ed3-b21e-380d43ea95a6
- What I've done
- 上午看[[学交]]，写[[数据库]]题
- 下午看[[产品开发]]
- 晚上看数电和毛概
- What I've thought #thoughts
- Mood
-
- [[产品开发]]
- ## Topic 7: Innovation
id:: 64857305-2925-485f-9aed-e15511b42149
collapsed:: true
- ### Definition of Innovation
Innovation is the process of creating something new that adds value to society. It can involve developing new products, services, processes, or business models. Innovation can be driven by a variety of factors, such as technological advances, changes in consumer preferences, or shifts in the competitive landscape.
- ### Types of Innovation
collapsed:: true
There are different types of innovation, including:
- Radical innovation: involves creating something entirely new that disrupts existing markets or creates new ones.
- Incremental innovation: involves making small improvements to existing products or processes.
- Disruptive innovation: involves creating a new product or service that initially serves a niche market but eventually disrupts the existing market.
- Sustaining innovation: involves making improvements to existing products or processes that help maintain a company's competitive position.
- ### Models of Innovation
collapsed:: true
Innovation can follow different models, such as:
- Linear model: involves a sequential process of research, development, and commercialization.
- Cyclical model: involves a continuous process of feedback and iteration.
- Open innovation: involves collaborating with external partners to develop new products or services.
- ### Characteristics of Innovative Companies
collapsed:: true
Innovative companies share certain characteristics, such as:
- A willingness to take risks and experiment with new ideas.
- A focus on customer needs and preferences.
- A culture that encourages creativity, collaboration, and learning.
- A commitment to continuous improvement and innovation.
- ### Strategies for Fostering a Culture of Innovation
collapsed:: true
Companies can foster a culture of innovation by:
- Providing resources, such as funding, time, and expertise, to support innovation initiatives.
- Encouraging collaboration and cross-functional teams.
- Rewarding creativity and risk-taking.
- Creating a supportive and inclusive work environment.
- ### Examples of Innovative Companies and Products
collapsed:: true
There are many examples of innovative companies and products, such as:
- Apple: known for its innovative products, such as the iPhone and iPad.
- Tesla: known for its innovative electric vehicles and renewable energy solutions.
- Sinclair C5: an innovative electric vehicle developed in the 1980s that was ahead of its time.
- ### Further Resources
The PDF provides links to further resources for learning about innovation, such as a video and short videos about new inventions. These resources can help individuals and companies stay up-to-date on the latest trends and developments in innovation and product development.
- ## Topic 7 PDF 概述
collapsed:: true
本 PDF 文件提供了创新和产品开发的概述。它涵盖了创新的基本原则、创新的类型、创新的模型以及什么使一个创新型公司。
- ## 涵盖的主题
collapsed:: true
- 创新的定义
- 创新的类型（根本性、渐进性、破坏性、维持性）
- 创新的模型（线性、循环、开放式创新）
- 创新型公司的特点
- 培养创新文化的策略
- 创新型公司和产品的例子
- 进一步阅读的资源（包括一个视频和关于新发明的短视频）
- ## 主要观点
collapsed:: true
- 创新是创造新的、对社会有价值的东西的过程。
- 有不同类型的创新，包括根本性、渐进性、破坏性和维持性。
- 创新可以遵循不同的模型，如线性、循环或开放式创新。
- 创新型公司共享某些特点，如愿意冒险、关注客户需求和鼓励实验和学习的文化。
- 公司可以通过提供资源、鼓励协作和奖励创造力来培养创新文化。
- 创新型公司和产品的例子包括苹果、特斯拉和辛克莱 C5 电动车。
- 进一步了解创新的资源包括一个视频和关于新发明的短视频。
- ### 创新的定义
创新是创造新的、对社会有价值的东西的过程。它可以涉及开发新产品、服务、流程或商业模式。创新可以由各种因素驱动，如技术进步、消费者偏好的变化或竞争格局的转变。
- ### 创新的类型
collapsed:: true
有不同类型的创新，包括：
- 根本性创新：涉及创造全新的东西，打破现有市场或创建新市场。
- 渐进性创新：涉及对现有产品或流程进行小幅改进。
- 破坏性创新：涉及创建一个最初服务于利基市场但最终打破现有市场的新产品或服务。
- 维持性创新：涉及对现有产品或流程进行改进，以帮助维持公司的竞争地位。
- ### 创新的模型
collapsed:: true
创新可以遵循不同的模型，如：
- 线性模型：涉及研究、开发和商业化的顺序过程。
- 循环模型：涉及反馈和迭代的持续过程。
- 开放式创新：涉及与外部合作伙伴合作开发新产品或服务。
- ### 创新型公司的特点
collapsed:: true
创新型公司共享某些特点，如：
- 愿意冒险和尝试新想法。
- 关注客户需求和偏好。
- 鼓励创造力、协作和学习的文化。
- 致力于持续改进和创新。
- ### 培养创新文化的策略
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公司可以通过以下方式培养创新文化：
- 提供资源，如资金、时间和专业知识，支持创新项目。
- 鼓励协作和跨职能团队。
- 奖励创造力和冒险精神。
- 创造一个支持性和包容性的工作环境。
- ### 创新型公司和产品的例子
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有许多创新型公司和产品的例子，如：
- 苹果：以其创新产品，如 iPhone 和 iPad 而闻名。
- 特斯拉：以其创新的电动汽车和可再生能源解决方案而闻名。
- 辛克莱 C5：一款在 1980 年代开发的创新电动车，领先于其时代。
- ### 进一步资源
本 PDF 文件提供了进一步了解创新的资源链接，如一个视频和关于新发明的短视频。这些资源可以帮助个人和公司了解创新和产品开发的最新趋势和发展。
- ## Topic 8: Digital Transofrmation and Digital Products
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- Digital Transformation
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- Terms
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- **Digitisation** is the process of converting information from analog to digital.
- **Digitalisation** is the process of using digitised information to make established ways of working simpler and more efficient.
- **Digital transformation** is the process of using digital technologies to create new — or modify existing — business processes, culture, and customer experiences to meet changing business and market requirements
- **Types** of Digital transformation
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- **Process Transformation** – aims to process such as data, analytics, AI, and any process that can work towards lowering costs and driving operational efficiency in the business.
- **Business Model Transformation** – aims to make fundamental changes in how a business or organization runs which can include personnel, processes, and technology.
- **Domain Transformation** - This area offers a great opportunity to move into a new domain or area that a business may not have explored before by acquiring new technologies.
- **Cultural/Organizational Transformation** -
This is about redefining mindsets, processes, capabilities and skills for a digital world. It’s about driving digital transformation forward through growth initiatives that are grounded in a new culture and way of thinking.
- **Guidelines** for a successful Digital
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Transformation
- Understand your technology
- Embrace Cultural Change
- Consider a new digital business model
- Digital upskilling
- Ensure Collaboration
- Top Management Support
- Digital Product
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- Characteristics
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- no physical form, exist only in the digital realm,
- _intangible_ items delivered _electronically_,
- anything that can be _downloaded_ and _used digitally_ can be considered a digital product,
- sold online or through brick-and-mortar retailers,
- can be easily updated or modified to keep up with changing technology and trends because they're intangible,
- often come with a license that allows a customer to use them in unlimited ways.
- Will everything become digital?
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- In practice, most products and experiences they are part of are _hybrid_.
- **Why** digital products?
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- Customer happiness is how you win in business. Modern customer expectations are being driven by largely digital technology and digital innovations.
- Low investment, (potentially) high returns
- More profitable than physical goods
- No inventory, shipping or rent hassle
- Automated delivery for passive income
- Serve a niche at scale
- Digital products offer unique ways to communicate directly with the customers.
- Digital Project Development
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- Phase 1. **Discovery**: the process of identifying the problem to be
solved, making sure the problem is worth solving, and
envisioning the solution to that problem.
- Phase 2. **Ideate**: The goal is to brainstorm possible solutions to the
problem identified in the discovery phase, creating a
strategy for how to build a product that will solve that
problem.
- Phase 3. **Test**: The testing phase is all about gathering data,
refining and improving your idea, and gathering more data
until you have a sharp idea. The testing phase should
involve at least four steps, known as the lean validation
process
- Phase 4. **Execute**: This stage aims to develop the“most
valuable player” namely the _Minimum Viable
Product (MVP)._
- Phase 5. **Launch**: Once your MVP is ready to go, it’s time to launch. A
digital product launch usually means putting the MVP
on the market and giving customers their first crack at
your solution.
- Phase 6. **Grow**: As refine the digital product into its final form, based on
the feedback from the MVP launch, it’s time to consider
the growth or scaling phase.
- Digital Project Management
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- Terms
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- **Digital Project Manager**: the glue that brings
together many facets of a successful digital product—
customers, design, engineering, operations, sales,
marketing, finance, compliance, legal, and more.
- **Project Team for Digital Products**:
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A project team for digital product is made up of:
- **Developers or engineers** – the people who will code, test and deploy the digital application that will be used by customers.
- **Experts in customer or user experience** who focus on how the product and associated services will be used by the customer, and who create the user interface, services and other interactions with the customers
- **Sales and marketing experts**, who will actually get customers to use the product.
- mistakes to avoid
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- Having the _wrong data_
- _Resistance_ from staff
- Underestimate _costs_
- A lack of _commitment_
- A lack of _skills_
- ## 主题 8
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- 数字化转型
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- 术语
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- **数字化** 是将信息从模拟形式转换为数字形式的过程。
- **数字化** 是利用数字化的信息使现有的工作方式更简单、更高效的过程。
- **数字化转型** 是利用数字技术来创建新的或修改现有的业务流程、文化和客户体验，以满足不断变化的业务和市场需求的过程。
- 数字化转型的**类型**
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- **流程转型** - 旨在通过数据、分析、人工智能以及任何可以降低成本并提高业务运营效率的流程来实现。
- **商业模式转型** - 旨在对企业或组织的运营进行根本性改变，包括人员、流程和技术。
- **领域转型** - 这个领域为企业提供了进入新领域或领域的机会，通过获取新技术来实现。
- **文化/组织转型** - 这涉及到重新定义数字世界中的思维方式、流程、能力和技能。通过以新的文化和思维方式为基础的增长举措，推动数字化转型的进程。
- 实现成功数字化转型的**指南**
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- 了解您的技术
- 接受文化变革
- 考虑新的数字商业模式
- 数字技能提升
- 确保合作
- 高层管理支持
- 数字产品
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- 特点
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- 无实体形态，只存在于数字领域
- *无形*产品通过*电子方式*交付
- 任何可以*下载*和*数字使用*的物品都可以被视为数字产品
- 在线销售或通过实体零售商销售
- 可以轻松更新或修改以跟上技术和趋势的变化，因为它们是无形的
- 通常附带许可证，允许客户以无限的方式使用它们
- 所有事物都会变成数字吗？
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- 实际上，大多数产品和体验都是*混合型*的。
- **为什么**选择数字产品？
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- 客户的幸福感是赢得业务的关键。现代客户的期望主要受数字技术和数字创新的推动。
- 投资低，（可能）回报高
- 比实体商品更具盈利能力
- 无需库存、物流或租金麻烦
- 自动化交付实现被动收入
- 以规模服务于细分市场
- 数字产品提供了与客户直接沟通的独特方式。
- 数字项目开发
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- 第一阶段 **发现**：识别待解决的问题，确保问题值得解决，并构想解决问题的方案。
- 第二阶段 **构思**：目标是对发现阶段确定的问题进行头脑风暴，创建构建解决该问题的产品的策略。
- 第三阶段 **测试**：测试阶段涉及收集数据、完善和改进您的想法，并收集更多数据，直到您拥有一个清晰的想法。测试阶段应包括至少四个步骤，即精益验证过程。
- 第四阶段 **执行**：这个阶段旨在开发最有价值的产品，即*最小可行产品 (MVP)*。
- 第五阶段 **发布**：一旦您的 MVP 准备就绪，就可以发布了。数字产品的发布通常意味着将 MVP 推向市场，让客户首次体验您的解决方案。
- 第六阶段 **增长**：在根据 MVP 发布的反馈对数字产品进行最终形态的改进时，是考虑增长或扩展阶段的时候。
- 数字项目管理
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- 术语
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- **数字项目经理**：他们将许多方面成功地融合在一起，包括客户、设计、工程、运营、销售、市场营销、财务、合规性、法律等等。
- **数字产品项目团队**：数字产品项目团队由以下成员组成：
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- **开发人员或工程师** - 负责编码、测试和部署将由客户使用的数字应用程序的人员。
- **用户体验专家** - 关注产品和相关服务如何被客户使用，并创建与客户的界面、服务和其他交互。
- **销售和市场营销专家** - 将实际使客户使用产品。
- 需要避免的错误
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- 使用*错误的数据*
- 员工*抵触*
- 低估*成本*
- 缺乏*承诺*
- 缺乏*技能*
- ## Topic 2: Project Management
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- **Define** what is project management
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- For all but the simplest products, product development involves many people completing many different tasks.
- is the activity of planning and coordinating resources and tasks to achieve these goals.
- Two Phases of **PM**
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- Project planning involves scheduling the project tasks and determining resource requirements. The project plan is first laid out during the concept development phase, although it is a dynamic entity and continues to evolve throughout the development process.
- Project execution, sometimes called project control, involves coordinating and facilitating the myriad tasks required to complete the project in the face of inevitable unanticipated events and the arrival of new information. Execution is just as important as planning; Many teams fail because they do not remain focused on their goals for the duration of the project.
- **Project Planning**: Understand and represent different tasks in projects
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- Sequential, Parallel, and Coupled Tasks
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- ![image.png](../assets/image_1686471430664_0.png)
- The Design Structure Matrix
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- **Definition**: A useful tool for representing and analysing
task dependencies is the design structure matrix
(DSM).
- Working:
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- A project task is assigned to a row and a corresponding column.
- The rows and columns are named and ordered identically, although generally only the rows list the complete names of the tasks. Each task is defined by a row of the matrix.
- We represent a task’s dependencies by placing marks in the columns to indicate the other tasks (columns) on which it depends.
- ![image.png](../assets/image_1686471695984_0.png)
- Reading across a row reveals all of the tasks whose output is required to perform the task corresponding to the row.
- Reading down a column reveals which tasks receive information from the task corresponding to the column.
- The diagonal cells are usually filled in with dots or the task labels, simply to separate the upper and lower triangles of the matrix and to facilitate tracing dependencies.
- Gantt Chart
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- Features:
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- Gantt charts show how the work is broken down into a set of activities
- They show the scheduling of these activities as a series of horizontal bands against a series of vertical lines representing dates
- ![image.png](../assets/image_1686471844454_0.png)
- They can be used to show dependencies between activities
- They can be used to measure progress on a project or compare planned production with actual production
- PERT Charts
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- PERT (program evaluation and review technique) charts explicitly represent both dependencies and timing, in effect combining some of the information contained in the DSM and Gantt chart.
- ![image.png](../assets/image_1686471880938_0.png)
- The Critical Path
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- The _dependencies_ among the tasks in a PERT chart, some of which may be arranged sequentially and some of which may be arranged in parallel, lead to the concept of a critical path.
- The _critical path_ is the longest chain of dependent events. This is the single sequence of tasks whose combined required times define the minimum possible completion time for the entire set of tasks.
- Undertake a baseline project plan
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- **Definition**: project plan is the roadmap for the remaining development effort. The plan is important in coordinating the remaining tasks and in estimating the required development resources and development time.
- Do project scheduling
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- Methods:
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- Contract Book
- Project Task List
- Team staffing & Organisation
- Project Schedule
- Project Budget
- Project Risk Plan
- Modifying the baseline plan
- Accelerate projects
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- Product development time is often the dominant concern in project planning and execution. There are a set of guidelines for accelerating product development projects.
- **Execute** projects
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- Smooth execution of even a well-planned project requires careful attention. Three problems of project execution are particularly important: 1. What mechanisms can be used to coordinate tasks? 2. How can project status be assessed? and 3. What actions can the team take to correct for undesirable deviations from the project plan?
- ## 主题 2
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- **定义** 什么是项目管理
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- 对于除了最简单的产品以外，产品开发涉及到许多人完成许多不同的任务。
- 项目管理是规划和协调资源和任务以实现这些目标的活动。
- **PM** 的两个阶段
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- 项目规划涉及安排项目任务并确定资源需求。项目计划首先在概念开发阶段制定，尽管它是一个动态实体，并在整个开发过程中不断演变。
- 项目执行，有时称为项目控制，涉及协调和促进完成项目所需的各种任务，面对必然发生的意外事件和新信息的到来。执行与规划同样重要；许多团队失败是因为在整个项目的持续时间内没有保持对目标的专注。
- **项目规划**：理解和代表项目中的不同任务
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- 顺序、并行和耦合任务
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- ![image.png](../assets/image_1686471430664_0.png)
- 设计结构矩阵
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- **定义**：用于表示和分析任务依赖关系的有用工具是设计结构矩阵（DSM）。
- 工作原理：
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- 一个项目任务分配给一行和相应的列。
- 行和列的命名和顺序相同，尽管通常只有行列出任务的完整名称。每个任务由矩阵的一行定义。
- 我们通过在列中放置标记来表示任务的依赖关系，以指示其所依赖的其他任务（列）。
- ![image.png](../assets/image_1686471695984_0.png)
- 横向读取一行可以看到所有需要完成与该行对应任务相关的任务的输出。
- 纵向读取一列可以看到哪些任务接收来自与该列对应的任务的信息。
- 对角线的单元格通常填充有点或任务标签，仅用于分隔矩阵的上三角形和下三角形，并便于追踪依赖关系。
- 甘特图
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- 特点：
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- 甘特图展示了工作如何被分解为一系列活动。
- 它们以一系列垂直线表示日期，以一系列水平带表示这些活动的调度。
- ![image.png](../assets/image_1686471844454_0.png)
- 它们可以用于显示活动之间的依赖关系。
- 它们可以用于衡量项目的进展或比较计划生产与实际生产情况。
- PERT 图
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- PERT（程序评估和审查技术）图明确表示依赖关系和时间，实际上结合了 DSM 和甘特图中包含的一些信息。
- ![image.png](../assets/image_1686471880938_0.png)
- 关键路径
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- PERT 图中任务之间的依赖关系，其中一些可能按顺序安排，一些可能并行安排，导致关键路径的概念。
- 关键路径是最长的一系列依赖事件。这是一系列任务的单一序列，其组合所需的时间定义了整个任务集合的最短可能完成时间。
- 进行基准项目计划
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- **定义**：项目计划是剩余开发工作的路线图。该计划在协调剩余任务和估计所需的开发资源和开发时间方面非常重要。
- 进行项目调度
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- 方法：
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- 合同书
- 项目任务列表
- 团队配备和组织
- 项目进度表
- 项目预算
- 项目风险计划
- 修改基准计划
- 加速项目
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- 产品开发时间通常是项目规划和执行中最主要的关注点。有一套加速产品开发项目的指南。
- **执行** 项目
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- 即使是一个精心计划的项目，顺利执行也需要仔细注意。项目执行的三个重要问题是：
1. 可以使用哪些机制来协调任务？
2. 如何评估项目状态？
3. 团队可以采取哪些行动来纠正与项目计划不符的不良偏差？
- ## Topic 3: Opportunity Identification
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- Opportunity, definition and types
- Definition:
- an idea for a new product
- It is a product description in an embryonic form
- a newly sensed need,
- a newly discovered technology,
- a rough match between a need and a possible solution
- It can be thought of as a hypothesis about how value might be created
- Type
- Ansoff’s growth matrix
![image.png](../assets/image_1686473366617_0.png)
- Market penetration
logseq.order-list-type:: number
- **Opportunities** can exist within a business’s existing
markets through increasing the volume of sales
of existing products
- market development
logseq.order-list-type:: number
- **Opportunities** are said to exist for a business’s
products through making them available to new
markets
- e.g. using existing products in new
geographical markets
- e.g. selling your existing products to a new age
group of customers
- product development
logseq.order-list-type:: number
- **Opportunities**: offering new or improved products to
logseq.order-list-type:: number
existing markets
- diversification
logseq.order-list-type:: number
- **Opportunities**: Moving into new markets, potentially with a base
logseq.order-list-type:: number
from your existing product knowledge or
diversification through acquisition of other
companies
- Risk
- The element of risk _increases the further the strategy moves away from known quantities_ the existing product and the existing market
- **Product development** (requiring, in effect, a new product) and market extension (a new market) involve a greater risk than market penetration
- **Diversification** (both new products and new markets) generally carries the greatest risk of all
- Opportunity identification process
- Establish **a charter**
- A **charter** articulates **the goals of the organisation**
(in relation to NPD) and establishes the **boundary
conditions for an innovation effort.** Charters can be
termed as mission statement for a new product.
- Generate and sense **many opportunities**
- Focus has to be both on **internal** and **external** sources of raw opportunities. Some of these are generated:
- Internally
- R&D department
- Externally
- customer
- competitive product
- sales forces
- collab. with universities
- investors
- distribution partners
- other partner companies
- Sense opportunities: Where do they come from?
- Passively
- Proactively
- Document **frustrations** and **complaints** that current **customers** experience with existing products
- Interview lead users, with attention devoted to
- **innovations** by these users and
- **modifications** these users may have
made to existing products
- **trends**
- Systematically gather suggestions from **current customers**
- **Competitors**
- **Transfer** emerging tech.
- **R&D**: _Research_ and _Development_
- **Definition**: to develop new knowledge and apply scientific or engineering knowledge to connect the knowledge in one field to that in others
- Roles:
- **Discovering and developing** new technologies Improving understanding of the technology in existing products
- **Improving and strengthening** understanding of technologies used in manufacturing
- **Understanding research results** from universities and other research institutions
- Areas:
- R&D for existing businesses
- R&D for new businesses
- R&D for exploratory research
- **Screen** opportunities
- Purpose:
- to eliminate any opportunities that are **unlikely to result in the creation of value**,
- to focus attention on the opportunities **worthy of further investigation**
- **not to** pick the _single best opportunity_!
- Approach:
- Web-based **surveys**
- Workshops with **multi-voting**: collaborative sessions or meetings where participants engage in a voting process to prioritize or make decisions on various options or ideas.
- Develop **promising opportunities**
- Details
- customer interviews,
- testing of existing products,
- concept generation,
- quick prototypes,
- estimates of market sizes and growth rates.
- Goal
- resolve the greatest uncertainty surrounding each one at the lowest cost in time and money.
- Select **exceptional opportunities**
- Goal
- select a few that warrant a significant investment in product development.
- Approach:**Real-Win-Worth-it**
- **Real**: Is the opportunity real?
- **Win**: Can you win with this opportunity?
- **Worth** It: Is it worth doing?
- **Reflect** on the result and process.
Ask the following questions:
- **How many** of the opportunities identified came from internal sources versus external sources?
- Did we consider **dozens or hundreds** of opportunities?
- Was the innovation charter **too narrowly focused**?
- Were our filtering criteria **biased**, or largely based on the best possible estimates of eventual product success?
- Are the resulting opportunities **exciting** to the team?
-
- ## 主题 3：机会识别
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- 机会的定义和类型
- 定义：
- 一种新产品的构想
- 它是一个初步形式的产品描述
- 新发现的需求
- 新发现的技术
- 需求和可能解决方案之间的初步匹配
- 可以将其视为关于如何创造价值的假设
- 类型
- 安索夫增长矩阵
![image.png](../assets/image_1686473366617_0.png)
- 市场渗透
- **机会**存在于企业现有市场中，通过增加现有产品的销售量来实现
- 市场开发
- 通过将产品面向新市场使得**机会**出现于企业的产品中
- 例如，在新的地理市场中使用现有产品
- 例如，将现有产品销售给新的客户群体
- 产品开发
- **机会**：向现有市场提供新产品或改进的产品
- 多元化
- **机会**：进入新市场，可能基于现有产品知识或通过收购其他公司进行多元化
- 风险
- 风险元素与策略偏离已知量（现有产品和现有市场）的程度成正比
- 与市场渗透相比，**产品开发**（实际上需要新产品）和市场扩展（新市场）涉及更大的风险
- **多元化**（新产品和新市场）通常带有最大的风险
- 机会识别过程
- 制定**章程**
- **章程**表达了组织（与 NPD 有关）的目标，并为创新工作设定了边界条件。章程可以被视为新产品的使命陈述。
- 生成和发现**众多机会**
- 需要关注内部和外部的机会来源。其中一些是通过以下方式生成的：
- 内部生成
- 研发部门
- 外部生成
- 客户
- 竞争产品
- 销售团队
- 与大学合作
- 投资者
- 分销合作伙伴
- 其他合作伙伴公司
- 感知机会：它们从何而来？
- 被动感知
- 主动感知
- 记录现有产品的当前客户所经历的**挫折**和**投诉**
- 与领先用户进行访谈，关注这些用户的**创新**以及他们对现有产品的**修改**
- 趋势
- 系统地收集来自**现有客户**的建议
- **竞争对手**
- **转移**新兴技术
- **研究与开发（R&D）**
- **定义**：开发新知识并应用科学或工程知识，将一领域的知识与其他领域相连接
- 角色：
- **发现和开发**新技术，提高对现有产品中技术的理解
- 改进和加强对制造中使用技术的理解
- 理解来自大学和其他研究机构的研究结果
- 领域：
- 现有业务的研发
- 新业务的研发
- 探索性研究的研发
- **筛选**机会
- 目的：
- 淘汰不太可能创造价值的机会
- 将注意力集中在值得进一步调查的机会上
- 不是选择*最佳机会*！
- 方法：
- 基于网络的调查
- 带有**多元投票**的工作坊：参与者进行投票以对各种选项或想法进行优先排序或决策的协作会议或会议。
- 开发**有前途的机会**
- 细节：
- 客户访谈
- 对现有产品进行测试
- 概念生成
- 快速原型
- 估计市场规模和增长率
- 目标：
- 以最低的时间和金钱成本解决每个机会周围最大的不确定性
- 选择**杰出的机会**
- 目标：
- 选择几个值得在产品开发中进行重大投资的机会
- 方法：**真实-胜利-值得**
- **真实**：机会是真实的吗？
- **胜利**：能够赢得这个机会吗？
- **值得**：值得去做吗？
- **反思**结果和过程。
提出以下问题：
- 机会识别中有多少机会来自内部来源和外部来源？
- 我们是否考虑了数十个还是数百个机会？
- 创新章程是否过于狭隘？
- 我们的筛选标准是否存在偏见，或者主要基于对最终产品成功的最佳估计？
- 结果的机会是否激发团队的激情？
- ## Topic 5: Identify Customer Needs (Preliminary of Concept Development)
id:: 64859157-dc41-453d-95ea-39e2280ea735
collapsed:: true
- **Market Research** for NPD (New Product Development)
collapsed:: true
- **Definition**: the function that links the consumer, customer
and public to the marketer through information
– information used
- **Goal**:
collapsed:: true
- to identify and define marketing opportunities and problems;
- to generate, refine and evaluate marketing actions;
- to monitor marketing performance;
- to improve understanding of the marketing process
- When to use it?
![image.png](../assets/image_1686475631266_0.png)
- **Gather** raw **data** from customers
logseq.order-list-type:: number
collapsed:: true
- **Interviews**: One or more development team members discuss with a single customer
- **Focus groups**: A moderator facilitates a two-hour discussion with a group of 8 or 12 customers
- **Observing** the product in use
- **Surveys**: direct mail or web-based questionnaires;
- **Interpret** the raw **data** in terms of customer needs
logseq.order-list-type:: number
collapsed:: true
- The data gathered in Step 1 then has to be used to
express the customers’ needs in terms of what the
product has to do, not in terms of how it might do
- Use positive, not negative phrasing
- Express the needs as attributes of the product
- Organize **the needs** into a hierarchy of needs
logseq.order-list-type:: number
collapsed:: true
- **Structure** the needs into
collapsed:: true
- **Must-haves** – *“I wont buy without”*
- **Delighters** – *“What an unexpected treat”*
- **Linear Satisfiers** – *“The more the merrier”*
- **Neutrals** – *“No big deal”*
- This is the *Kano* classification
- Consumer needs can be very **elusive**
- **Intuitions** are often wrong
- Establish the **relative importance** of the needs
logseq.order-list-type:: number
collapsed:: true
- Measuring **preference** is central to market research
- **Reflect** on the results and the process
logseq.order-list-type:: number
collapsed:: true
- Questions to ask:
collapsed:: true
- Have we interacted with all **important customers** in
our target market?
- Can we see the **latent needs** of customers beyond
our current product range?
- Can we **further involve** any of the customers in our
product development?
- Did we involve the **right people** in our organisation?
- Can we **improve** our process?
- ## 第五主题：识别顾客需求（概念开发的初步阶段）
collapsed:: true
- **NPD（新产品开发）的市场调研**
- **定义**：通过信息将消费者、顾客和公众与营销人员联系起来的功能
- **目标**：
- 识别和定义市场营销机会和问题；
- 生成、改进和评估市场营销行动；
- 监测市场营销绩效；
- 提高对市场营销过程的理解
- 何时使用？
![image.png](../assets/image_1686475631266_0.png)
- 从顾客那里收集原始数据
logseq.order-list-type:: number
- **面谈**：一个或多个开发团队成员与单个顾客讨论
- **焦点小组**：一个主持人与8至12位顾客进行两小时的讨论
- 观察产品的使用情况
- **调查**：直邮或基于网络的问卷调查；
- 将原始数据解读为顾客需求
logseq.order-list-type:: number
- 然后，必须使用在步骤1中收集到的数据来表达顾客的需求，而不是以产品如何满足需求的方式来表达
- 使用积极的、而非消极的措辞
- 将需求表达为产品的属性
- 将需求组织成层次结构
logseq.order-list-type:: number
- 将需求分为以下部分：
- **必备项** - “如果没有这个，我就不会购买”
- **惊喜项** - “真是个意外的惊喜”
- **线性满足项** - “越多越好”
- **中立项** - “无关紧要”
- 这是*Kano*分类
- 消费者需求可能非常难以捉摸
- 直觉经常是错误的
- 确定需求的**相对重要性**
logseq.order-list-type:: number
- 测量**偏好**是市场调研的核心
- **反思**结果和过程
logseq.order-list-type:: number
- 需要问的问题：
- 我们是否与目标市场中的所有**重要顾客**进行了互动？
- 我们能否看到顾客在我们当前产品范围之外的**潜在需求**？
- 我们能否在产品开发中**进一步吸引**任何顾客的参与？
- 我们是否在组织中吸引了**合适的人员**？
- 我们能否改进我们的流程？

7
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- Daily reflection [[Daily reflections]]
- What I've done
- 早上看java
- 下午看java，考试
- 晚上看毛概，背不动，做题，和chatgpt聊政治，看数据库Relational algebra，看了一点数电
- What I've thought #thoughts
- Mood

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- #to_be_deleted [[New computer 2023]]
- 灵刃 15 2023 32 + 1T
id:: 64224ad1-0983-437e-89bc-779520a862aa
- | 日期 | 价格 |
| ------------------ | ----- |
| [[Mar 28th, 2023]] | 20999 |
| [[Apr 2nd, 2023]] | 21499 |
- 灵刃 15 2023 16 + 1T
- | 日期 | 价格 |
| ----------------- | ----- |
| [[Apr 2nd, 2023]] | 19999 |

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