Find Minimum in Rotated Sorted Array

一般的二分查找都是low = mid - 1或者high = mid + 1之类来避免重复，但此题必须要比较两个数值，故而要low=mid或者high = mid。

public class Solution {
    public int findMin(int[] num) {
        if(num == null || num.length == 0) return 0;
        int low = 0, high = num.length - 1;
        while(low < high) {
            if(low == high - 1 && num[low] > num[high]) return num[high];
            int mid = low + (high - low) / 2;
            if(num[low] < num[mid]) {
                low = mid;
            } else {
                high = mid;   
            }
        }
        return num[0];
    }
}

另外一个简洁一点的解法是： 

public class Solution {
    public int findMin(int[] num) {
        int low = 0, high = num.length - 1;
        while(low < high && num[low] > num[high]) {
            int mid = low + (high - low) / 2; 
            if(num[low] <= num[mid]) { //因为low可能等于mid，所以要考虑等于的情况
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return num[low];
    }
}