> For the complete documentation index, see [llms.txt](https://protegejj.gitbook.io/oj-practices/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://protegejj.gitbook.io/oj-practices/chapter1/binary-search/81-search-in-rotated-sorted-array-ii.md).

# 81 Search in Rotated Sorted Array II

## 81. [Search in Rotated Sorted Array II](https://leetcode.com/problems/search-in-rotated-sorted-array-ii/description/)

## 1. Question

Follow up for "Search in Rotated Sorted Array":\
What ifduplicatesare allowed?

> Would this affect the run-time complexity? How and why?

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

(i.e.,`0 1 2 4 5 6 7`might become`4 5 6 7 0 1 2`).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

## 2. Implementation

**(1) Binary Search**

思路: 这题和33题的区别在于数组存在重复，但基本思路还是一样，由于二分查找的运用前提是数组要有序的，所以分成4种情况:

(1) nums\[mid] 等于 target，找到target，直接返回true

(2) nums\[mid] < nums\[end]， \[mid, end]区间是有序的

* 如果nums\[mid] < target <= nums\[end], 在\[mid  + 1, end]区间继续做二分搜索
* 否则在\[start, mid - 1]做二分搜索

(3) nums\[mid] > nums\[end], 由于数组时rotate一次，所以\[start, mid]是有序的

* 如果nums\[start <= target < nums\[mid]， 在\[start, mid - 1]区间做二分搜索
* 否则在\[mid + 1, end]二分搜索

(4) nums\[mid]等于nums\[end]，此时无法判断方向，end往左移一步

```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, mid = 0;

        while (start + 1 < end) {
            mid = start + (end - start) / 2;

            if (nums[mid] == target) {
                return true;
            }
            else if (nums[mid] < nums[end]) {
                if (nums[mid] < target && target <= nums[end]) {
                    start = mid + 1;
                }
                else {
                    end = mid - 1;
                }
            }
            else if (nums[mid] > nums[end]) {
                if (nums[start] <= target && target < nums[mid]) {
                    end = mid - 1;
                }
                else {
                    start = mid + 1;
                }
            }
            else {
                --end;
            }
        }

        return nums[start] == target || nums[end] == target;
    }
}
```

## 3. Time & Space Complexity

时间复杂度O(logn)，空间复杂度O(1)