> For the complete documentation index, see [llms.txt](https://rachel2011.gitbook.io/leetcode_cc150/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://rachel2011.gitbook.io/leetcode_cc150/binary-search/find-minimum-in-rotated-sorted-array-ii.md).

# Find Minimum in Rotated Sorted Array II

Follow up for "Find Minimum in Rotated Sorted Array":\
What if duplicates are 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`).

Find the minimum element.

The array may contain duplicates.

寻找旋转有序重复数组的最小值是对Find Minimum in Rotated Sorted Array 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end减一，略过与mid相同的数字，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)

```
class Solution(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """  
        start, end = 0, len(nums)-1
        while start + 1 < end:
            mid = (start+end)/2
            #if mid equals to end, that means it's fine to remove end,the smallest element won't be removed
            if nums[mid] == nums[end]:
                end -= 1
            elif nums[mid]>nums[end]:
                #minV = min(minV, nums[start])
                start = mid
            else:
                #minV = min(minV, nums[mid])
                end = mid
        return min(nums[start], nums[end])
```