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 a sorted array 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.
153的改进版，可能会有重复的数字出现，
所以需要单独讨论等于的情况，等于时上界下移，
这样在数字全部相同时会退化为O（n）复杂度
代码语言：javascript复制class Solution {
public:
int findMin(vector<int>& nums) {
int l = 0, r = nums.size()-1;
while(l<r)
{
int mid = (l r) >> 1;
if(nums[mid] > nums[r])
l = mid 1;
else if(nums[mid] < nums[r])
r = mid;
else
r--;
}
return nums[l];
}
}; 
