---
description: Binary Search
---

# LeetCode 33. Search in Rotated Sorted Array

There is an integer array `nums` sorted in ascending order (with **distinct** values).

Prior to being passed to your function, `nums` is **rotated** at an unknown pivot index `k` (`0 <= 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 rotation and an integer `target`, return _the index of_ `target` _if it is in_ `nums`_, or_ `-1` _if it is not in_ `nums`.

**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`
* `-10^4 <= nums[i] <= 10^4`
* All values of `nums` are **unique**.
* `nums` is guaranteed to be rotated at some pivot.
* `-10^4 <= target <= 10^4`

&#x20;**Follow up:** Can you achieve this in `O(log n)` time complexity?

## Solution

[English Version in Youtube](https://youtu.be/_0_8c6kZlxM)

[中文版解答Youtube Link](https://youtu.be/gSevuT3fvZU)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1154y187P5/)

```
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int i = 0;
        int j = nums.size() - 1;
        
        while (i <= j) {
            int mid = (i + j) / 2;
            if (nums[mid] == target) return mid;
            if (nums[mid] < nums[i]) {
                if (nums[mid] < target && nums[j] >= target) i = mid+1;
                else j = mid-1;
            } else {
                if (nums[mid] > target && nums[i] <= target) j = mid-1;
                else i = mid+1;
            }
        }
        
        return -1;
    }
};
```
