154. Find Minimum in Rotated Sorted Array II

class Solution {
    public int findMin(int[] nums) {
      int start = 0;
      int end = nums.length - 1;
      while(start < end){
        int mid = start + (end - start) / 2;
        if(nums[mid] < nums[end]){
          end = mid;
        }else if( nums[mid] > nums[end]){
          // 678901
          start = mid + 1;
        }else{
          end--;
        }
      }
      return nums[start];
        
    }
}

 

和Search in Rotated Sorted Array II这题换汤不换药。同样当A[mid] = A[end]时，无法判断min究竟在左边还是右边。

 

3 1 2 3 3 3 3 
3 3 3 3 1 2 3

 

但可以肯定的是可以排除A[end]：因为即使min = A[end]，由于A[end] = A[mid]，排除A[end]并没有让min丢失。所以增加的条件是：

 

A[mid] = A[end]：搜索A[start : end-1]

 

 

 

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 duplicates.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

• This is a follow up problem to Find Minimum in Rotated Sorted Array.
• Would allow duplicates affect the run-time complexity? How and why?