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.

解法1：

//自己想的，遇到首尾和mid相等的，檢驗mid是左邊還是在右邊，檢驗時只能挨個查詢。
    int findMin1(vector<int>& nums) {
        int n=nums.size();
        if(n==1) return nums[0];
        int first=0,last=n-1;
        while(first<last){
            int mid=(first+last)/2;
            if
 
(nums[last]<nums[mid]) first=mid+1;
            else{
                if(nums[first]==nums[mid]&&nums[mid]==nums[last]){
                    int l=mid;
                    while(l<last&&nums[++l]==nums[last]);//檢驗mid是否在右邊
                    if(l==last) last=mid;
                    else
 
 first=mid+1;
                }
                else last=mid;
            }
          }
        return nums[first];
    }

解法2：思路更加清晰，易懂。

//按照陳老師的優化了思路，直接把開始的first(nums[first]==nums[last])的重複的去掉。  
//後面和沒重複的一樣了！
    int findMin(vector<int>& nums) {
        int n=nums.size();
        if(n==1) return nums[0];
        int first=0,last=n-1;
        while(first<last){
             while(nums[first]==nums[last]&&first<last){
                    first++;
             } 
             int mid=(first+last)/2;
             if(nums[last]<nums[mid]) first=mid+1;
             else  last=mid;
         }
         return nums[first];
    }