Find Minimum in Rotated Sorted Array leetcode java

题目：

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.

解题思路：

首先假设一个sorted没有rotated的数组[1,2,3]，假设我们通过一个pivot把这个数组rotate，那么结果可能为：[2,3,1], [3,1,2], 可以发现：num[low]永远大于（或等于）num[high]。因为你之前是sorted的数组，你在一个sorted的数组找了一个pivot进行rotate，那么比如pivot后面的值都大于pivot之前的值。所以依据这个发现，以及二分法查找。我们可以根据以下判断来解题。num[mid]有两种可能性，如果num[mid] > num[high]，证明num[mid]在rotated后的那个区间内，这个区间我们刚才已知都大于pivot之前的值，所以最小值就在low=mid+1那个区间内。另一种可能，num[mid] <= num[high]，那么我们刚才可以看出来这种可能性说明mid~high以及是排好序的，那么最小值在high=mid这个区间内（mid可能是最小值）。依据此判断可以找到最小值。

代码如下：

1     public int findMin(int[] num) {
2         int low = 0, high = num.length - 1;
3         while (low < high && num[low] >= num[high]) {
4             int mid = (low + high) / 2;
5             if (num[mid] > num[high])
6                 low = mid + 1;
7             else
8                 high = mid;
9         }
10         return num[low];
11     }

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