107 lines
1.8 KiB
Markdown
107 lines
1.8 KiB
Markdown
# Leetcode Binary-Search
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#### 2022-07-09 09:34
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> ##### Algorithms:
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>
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> #algorithm #binary_search
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>
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> ##### Data structures:
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>
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> #DS #array
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>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty-easy
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>
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> ##### Additional tags:
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>
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> #leetcode #CS_list_need_practicing
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/binary-search/)
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---
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### Problem
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Given an array of integers `nums` which is sorted in ascending order, and an integer `target`, write a function to search `target` in `nums`. If `target` exists, then return its index. Otherwise, return `-1`.
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You must write an algorithm with `O(log n)` runtime complexity.
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#### Examples
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**Example 1:**
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**Input:** nums = [-1,0,3,5,9,12], target = 9
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**Output:** 4
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**Explanation:** 9 exists in nums and its index is 4
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**Example 2:**
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**Input:** nums = [-1,0,3,5,9,12], target = 2
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**Output:** -1
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**Explanation:** 2 does not exist in nums so return -1
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#### Constraints
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- `1 <= nums.length <= 104`
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- `-104 < nums[i], target < 104`
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- All the integers in `nums` are **unique**.
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- `nums` is sorted in ascending order.
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### Thoughts
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> [!summary]
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> This is a #binary_search problem.
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Key takeout:
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```
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int r = nums.size() - 1;
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```
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make sure r is never OOB (l == r && r = array.size())
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### Solution
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==#TODO: write in recursion==
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iteration
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```cpp
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class Solution {
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public:
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int search(vector<int> &nums, int target) {
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// Why - 1?
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// Make sure mid is never OOB (l == r && r = size)
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int r = nums.size() - 1;
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int l = 0;
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int mid;
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int val;
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do {
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// l + ( r - l ) / 2 or (l + r) / 2 Both ok
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mid = (l + r) / 2;
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val = nums[mid];
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if (val == target) {
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return mid;
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} else if (val < target) {
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l = mid + 1;
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} else {
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r = mid - 1;
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}
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} while (l <= r);
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return -1;
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}
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};
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```
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