Add Date 2014-10-15
Find Minimum in Rotated Sorted Array
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.
这个很简单，因为没有重复数字，数组本质上还是有序的，用类似二分查找的方法复杂度O(log n)。code 也没几行，直接看 code 吧，记得考虑一下整个数组没有 rotated 的情况。
1 class Solution { 2 public: 3 int findMin(vector<int> &num, int pre, int post) { 4 if(pre == post) 5 return num[pre]; 6 if(num[pre] < num[post]) 7 return num[pre]; 8 int mid = (pre+post)/2; 9 if(num[mid] >= num[pre]) 10 return findMin(num, mid+1, post); 11 if(num[mid] < num[post]) 12 return findMin(num, pre, mid); 13 } 14 int findMin(vector<int> &num) { 15 int len = num.size(); 16 return findMin(num, 0, len-1); 17 } 18 };
明天继续刷LeetCode。。
【LeetCode】Find Minimum in Rotated Sorted Array 在旋转数组中找最小数
原文：http://www.cnblogs.com/shirley130912/p/4029863.html