Find Minimum in Rotated Sorted Array II

Description:

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.

Code:

 bool flag = true;
 int findMin(vector<int>& nums) {
    if (nums.size()==0){
          flag = false;
      return 0;
        }
       int left = 0;
       int right = nums.size()-1;
       while (left < right)
       {
           int mid = (left + right)/2;
           if (nums[mid] < nums[right])
                right = mid;
           else if (nums[mid] > nums[right])
                left = mid+1;
           else
           {
               //此时不能判最小值在左边还右边,遍历判断
                int result = nums[left];
                for (int i = left+1; i <= right; i++)
                {
                    if ( nums[i] < result)
                         result = nums[i];
                }
                return result; 
           }
       }
       return nums[left];
    }
 
原文地址:https://www.cnblogs.com/happygirl-zjj/p/4592202.html