LeetCode之Find Minimum in Rotated Sorted Array II

这道题目和上一道题目描述差不多，唯一多了的陷阱是可能会出现重复的数字，因此判断low 和high的值需要谨慎点，题目描述如下：

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.

The array may contain duplicates.

具体的java代码如下：

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