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：如果中间节点的值最大，则取后半部分，如果中间节点的值最小，则取前半部分。
C++实现代码如下：（采用二分法的方法，注意边界的处理）
#include<iostream> #include<vector> using namespace std; class Solution { public: int findMin(vector<int> &num) { if(num.empty()) return 0; int s=0; int t=num.size()-1; if(num[s]<num[t]) return num[s]; while(s<t) { int mid=(s+t)/2; if(num[mid]<num[t]) t=mid; else s=mid+1; } return num[s]; } }; int main() { Solution s; vector<int> num={4,5,6,7,8,9,0,1,2,3}; cout<<s.findMin(num)<<endl; }
Find Minimum in Rotated Sorted Array
原文：http://www.cnblogs.com/wuchanming/p/4102634.html