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.
分析：Seearch in rotated sorted array的变种，关键在于确定binary search的左右边界。代码如下：
class Solution { public: int findMin(vector<int> &num) { int n = num.size(); if(n == 1) return num[0]; if(n == 2) return min(num[0], num[1]); if(num[0] < num[n-1]) return num[0]; int l = 0, r = n-1; while(l <= r){ int mid = (l + r)/2; if(mid == n-1 && num[mid] < num[mid-1] || num[mid] < num[mid-1] && num[mid] < num[mid+1]) return num[mid]; if(num[mid] >= num[0]) l = mid + 1; else r = mid - 1; } } };
Find Minimum in Rotated Sorted Array
原文：http://www.cnblogs.com/Kai-Xing/p/4059121.html