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.

 Notice

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]);    }}