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 are arranged 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 tonight without alerting the police.

     这道题和之前它的基础版最大的不同就是所有house围成了一个圈。之前我们只需要考虑这个house的前一个是否被rob与否。现在呢就有了特殊情况。

     第一个和最后一个house不能同时被rob。

     最为简单的方法就是call两次第一次的算法。（一个assume第一个house被rob，一个则assume最后一个house被rob。）

     因为这里有nums.length-2，因此special case里面应该有nums.length==1的时候的单独的算法。

     代码如下。~

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);
         
    }
}