154. Find Minimum in Rotated Sorted Array II leetcode binary search

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.
注意：该数组中包含重复的元素
int findMin(vector<int>& nums) {
int low = 0;
int high = nums.size() - 1;

while(low < high)
{
if(nums[low] < nums[high])
return nums[low];

//进入到以下部分说明不能满足上述的if条件，即以下部分nums[low] > nums[high]
int mid = low + (high - low) / 2;
if(nums[mid] > nums[low])
low = mid + 1;
else if(nums[mid] < nums[low])
high = mid; //此处没有high = mid + 1 是因为nums[mid] < nums[high],那么mid处有可能是最先的数字
else
low++; //多加了一句，即当nums[mid] == nums[low]的时候，因为数组中有重复的元素，所以不清楚最小的元素出现在哪部分，是出现在low~mid之间还是mid~high之间，所以只将low++
}

return nums[low];
}