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.
You may assume no duplicate exists in the array.
思路：采取二分方法,将中间的值与边界进行相比较。
1 class Solution { 2 public: 3 int findMin(vector<int>& nums) { 4 int left = 0,right = nums.size()-1; 5 while(left<right){ 6 int middle = (left + right)/2; 7 if(nums[middle]>=nums[right]) 8 left = middle+1; 9 else 10 right = middle; 11 } 12 return nums[left]; 13 14 } 15 };
转载于:https://www.cnblogs.com/GoFly/p/5754516.html
