# LeetCode 81. Search in Rotated Sorted Array II

There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).

Before 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, `[0,1,2,4,4,4,5,6,6,7]` might be rotated at pivot index `5` and become `[4,5,6,6,7,0,1,2,4,4]`.

Given the array `nums` **after** the rotation and an integer `target`, return `true` *if* `target` *is in* `nums`*, or* `false` *if it is not in* `nums`*.*

**Example 1:**

```
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
```

**Example 2:**

```
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
```

**Constraints:**

* `1 <= nums.length <= 5000`
* `-104 <= nums[i] <= 104`
* `nums` is guaranteed to be rotated at some pivot.
* `-104 <= target <= 104`

&#x20;**Follow up:** This problem is the same as [Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/description/), where `nums` may contain **duplicates**. Would this affect the runtime complexity? How and why?

## Solution

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

[中文版解答Youtube Link](https://youtu.be/tBfnbjEw6RQ)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1g54y187cX/)

```
class Solution {
public:
    bool 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 true;
            
            if (nums[mid] < nums[i]) {
                if (nums[mid] < target && nums[j] >= target) i = mid + 1;
                else j = mid - 1;
            } else if (nums[mid] > nums[i]) {
                if (nums[mid] > target && nums[i] <= target) j = mid - 1;
                else i = mid + 1;
            } else {
                i++;
            }
        }
        
        return false;
    }
};
```