LeetCode 153.Find Minimum in Rotated Sorted Array(M)(P)

题目：

Suppose an array sorted in ascending order 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：

1.题目中指出sorted array，find target使用二分法。

2.分为以下三种情况

（1）数组只有一个元素：nums[0]即为最小值。

（2）数组不是rotated sorted array，退化为sorted array：nums[0]即为最小值。

（3）数组是rotated sorted array：nums[start] > nums[end]：

a:nums[mid]大于nums[start].

b:nums[mid]小于nums[start].

代码：

public class Solution {
    public int findMin(int[] nums) {
        int start = 0,end = nums.length-1,mid = 0;
           //length equals 1；
        if(nums.length == 1){
            return nums[0];
        }
           //is rotated sorted array;
        while(nums[start] > nums[end] && start + 1 < end){
            mid = start + (end - start)/2;
            if(nums[mid] > nums[start]){
                start = mid;
            }else if(nums[mid] < nums[start]){
                end = mid;
            }
        }
           //is not rotated sorted array;
        if(nums[start] < nums[end]){
            return nums[start];
        }
        return nums[end];
    }
}