After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place arearranged in a circle.That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the PRevious street.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonightwithout alerting the police.

這道題和之前它的基礎版最大的不同就是所有house圍成了一個圈。之前我們只需要考慮這個house的前一個是否被rob與否。現在呢就有了特殊情況。

第一個和最后一個house不能同時被rob。

最為簡單的方法就是call兩次第一次的算法。（一個assume第一個house被rob，一個則assume最后一個house被rob。）

因為這里有nums.length-2，因此special case里面應該有nums.length==1的時候的單獨的算法。

代碼如下。~

public class Solution {    public int rob(int[] nums) {        if(nums.length==0||nums==null){            return 0;        }        return Math.max(rob1(nums,0,nums.length-2),rob1(nums,1,nums.length-1));                    }        private int rob1(int[]nums, int start,int end){         if(nums.length==0||nums==null){            return 0;         }         if(nums.length==1){             return nums[0];         }         int prerob=0;         int predontrob=0;         for(int i=start;i<end+1;i++){             int curr=predontrob+nums[i];             int dontcurr=Math.max(prerob,predontrob);             prerob=curr;             predontrob=dontcurr;         }         return Math.max(prerob,predontrob);             }}