154. Find Minimum in Rotated Sor

ollow 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

仍然是在旋转有序数组中查找最小值，但是此时元素允许重复。有可能遇到如图所示的情况。

此时无法继续判断，只能退回到线性遍历的方式。

//顺序遍历辅助函数
int help(vector<int> &num,int start,int end){
        int min_v = num[start];
        for(int i=start+1;i<=end;i++){
            min_v = min(min_v,num[i]);
        }
        return min_v;
    }
    
    int findMin(vector<int> &num) {
        int start=0,end=num.size()-1;
        
        while (start<end) {
            if (num[start]<num[end])
                return num[start];
            
            int mid = (start+end)/2;
            
            if(num[mid]==num[start] && num[mid]==num[end]){
                return help(num,start,end);
            }
            
            if (num[mid]>=num[start]) {
                start = mid+1;
            } else {
                end = mid;
            }
        }
        
        return num[start];
    }