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.
Example
Given [4, 5, 6, 7, 0, 1, 2] return 0
OOXX二分法（find the first X）。特征：相对最后一个数的大小。O：>nums[end]； X: <=nums[end]。
切记要和最后一个数比，不可以和第一个数比，因为和第一个数比在没有rotate过的特殊情况里会找空。
public class Solution { /* * @param nums: a rotated sorted array * @return: the minimum number in the array */ public int findMin(int[] nums) { // write your code here int start = 0; int end = nums.length - 1; int cmpTarget = nums[end]; while (start + 1 < end){ int mid = start + (end - start) / 2; if (nums[mid] > cmpTarget){ start = mid; } else { end = mid; } } return Math.min(nums[start], nums[end]); } }
lintcode159- Find Minimum in Rotated Sorted Array- medium
原文：http://www.cnblogs.com/jasminemzy/p/7580058.html