之前一步一步存的方法略复杂啊，其实直接二分扔就可以咯
这里直接给个II的解法，和I多个冗余挪位的判断
public class Solution { public int findMin(int[] nums) { int min = nums[0]; int l=0, h = nums.length-1; while(l<=h){ int m = (l+h)/2; if(nums[m]<nums[h]){ h =m-1; }else if(nums[m]>nums[h]){ l = m+1; }else{ h--; } min = Math.min(min,nums[m]); } return min; } }
Find Minimum in Rotated Sorted Array I, II
原文：http://www.cnblogs.com/jiajiaxingxing/p/4438939.html