107 lines
1.9 KiB
Markdown
107 lines
1.9 KiB
Markdown
# Leetcode Search-Insert-Position
|
|
|
|
#### 2022-07-09 10:25
|
|
|
|
> ##### Algorithms:
|
|
>
|
|
> #algorithm #binary_search
|
|
>
|
|
> ##### Data structures:
|
|
>
|
|
> #DS #array
|
|
>
|
|
> ##### Difficulty:
|
|
>
|
|
> #coding_problems #difficulty_easy
|
|
>
|
|
> ##### Additional tags:
|
|
>
|
|
> #leetcode
|
|
>
|
|
> ##### Revisions:
|
|
>
|
|
> N/A
|
|
|
|
##### Related topics:
|
|
|
|
##### Links:
|
|
|
|
- [Link to problem](https://leetcode.com/problems/binary-search/)
|
|
|
|
***
|
|
|
|
### Problem
|
|
|
|
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`.
|
|
|
|
You must write an algorithm with `O(log n)` runtime complexity.
|
|
|
|
#### Examples
|
|
|
|
**Example 1:**
|
|
|
|
**Input:** nums = [-1,0,3,5,9,12], target = 9
|
|
**Output:** 4
|
|
**Explanation:** 9 exists in nums and its index is 4
|
|
|
|
**Example 2:**
|
|
|
|
**Input:** nums = [-1,0,3,5,9,12], target = 2
|
|
**Output:** -1
|
|
**Explanation:** 2 does not exist in nums so return -1
|
|
|
|
#### Constraints
|
|
|
|
- `1 <= nums.length <= 104`
|
|
- `-104 < nums[i], target < 104`
|
|
- All the integers in `nums` are **unique**.
|
|
- `nums` is sorted in ascending order.
|
|
|
|
### Thoughts
|
|
|
|
> [!summary]
|
|
> This is a #binary_search problem.
|
|
|
|
similar to [[Leetcode First-Bad-Version]], this relies on the converging nature of Bi-search
|
|
|
|
if the value is found, simply return it
|
|
|
|
if not found, at the end, position `l - 1` should be smaller than val.
|
|
|
|
and position `l` should be bigger, which is the position for the answer:
|
|
|
|
```
|
|
If not, return the index where it would be if it were inserted in order.
|
|
```
|
|
|
|
### Solution
|
|
|
|
Iteration
|
|
|
|
```cpp
|
|
class Solution {
|
|
public:
|
|
int searchInsert(vector<int>& nums, int target) {
|
|
// variant of bi-search
|
|
int r = nums.size() - 1;
|
|
int l = 0;
|
|
int mid, val;
|
|
|
|
do {
|
|
mid = l + (r - l) / 2;
|
|
val = nums[mid];
|
|
|
|
if (val == target) {
|
|
return mid;
|
|
} else if (val < target) {
|
|
l = mid + 1;
|
|
} else {
|
|
r = mid - 1;
|
|
}
|
|
} while (l <= r);
|
|
|
|
return l;
|
|
}
|
|
};
|
|
```
|