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
class Solution { public: int findMin(vector<int>& nums) { int len=nums.size(); if(len==1) return nums[0]; int low=0,high=len-1; //有重复的数字从mid 和high 比较来辨别 寻找的位于左半部分还是右半部分,没有的话用 mid和low位置比较划分 while(low<=high){ int mid=low+(high-low)/2; if(nums[mid]<nums[high]) { high=mid; }else if(nums[mid]>nums[high]){ low=mid+1; }else{ high--; } } return nums[low]; } };