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

# Find Minimum in Rotated Sorted Array II

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]`).

Find the minimum element.

The array may contain duplicates.

**Example 1:**

```
Input: [1,3,5]
Output: 1
```

**Example 2:**

```
Input: [2,2,2,0,1]
Output: 0
```

**Note:**

* This is a follow up problem to [Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/).
* Would allow duplicates affect the run-time complexity? How and why?

寻找旋转有序重复数组的最小值是对之前问题的延伸，当数组中存在大量的重复数字时，就会破坏二分查找法的机制，我们无法取得O(lgn)的时间复杂度，又将会回到简单粗暴的O(n)，比如如下两种情况：

{2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2} 和 {2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2}， 我们发现，当第一个数字和最后一个数字，还有中间那个数字全部相等的时候，二分查找法就崩溃了，因为它无法判断到底该去左半边还是右半边。这种情况下，我们将左指针右移一位，略过一个相同数字，这对结果不会产生影响，因为我们只是去掉了一个相同的，然后对剩余的部分继续用二分查找法，在最坏的情况下，比如数组所有元素都相同，时间复杂度会升到O(n)，

对比之前rotation array 有重复的情况

```java
class Solution {
    public int findMin(int[] nums) {
        if(nums == null | nums.length == 0){
            return -1;
        }
        
        
        int start = 0, end = nums.length - 1;
        
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            
            if(nums[mid] < nums[end]){
                end = mid;
            }
            else if(nums[mid] > nums[end]){
                start = mid;
            }
            //因为要找最小值，所以尽量从后往前缩进，end--跳过重复值
            else{
                end--;
            }
        }
        
        if(nums[start]<= nums[end]){
            return nums[start];
        }else{
            return nums[end];
        }
        
    }
}
```