> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/binary-search/81.search-in-rotated-sorted-array-ii.md).

# 81.Search-in-Rotated-Sorted-Array-II

## 81. Search in Rotated Sorted Array II

## 题目地址

<https://leetcode.com/problems/search-in-rotated-sorted-array-ii/>

## 题目描述

```
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

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
Follow up:

This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
Would this affect the run-time complexity? How and why?
```

## 代码

### Approach #1

<https://segmentfault.com/a/1190000016825704>

```java
class Solution {
    public boolean search(int[] nums, int target) {
        if (nums == null || nums.length == 0) return false;

        int start = 0, end = nums.length - 1;
        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) return true;
            if (nums[mid] > nums[start]) {
                if (target < nums[mid] && target >= nums[start]) {
                    end = mid;
                } else {
                    start = mid + 1;
                }
            } else if (nums[mid] < nums[start]) {
                if (target > nums[mid] && target < nums[start]) {
                    start = mid + 1;
                } else {
                    end = mid;
                }
            } else {
                start++;
            }
        }
        return false;
    }
}
```
