213. House Robber II

Note: This is an extension of House Robber.

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.

和之前的house robber差不多，只能隔着房间偷东西，要是两个连着的房间偷了就会触发警报，现在房间头尾连起来了。问如何解决问题。

解题点，头尾相连，

因为头尾相连所以头尾这两个房子只能偷一个，这样就形成了两种可能，

0到n-2之间的和1到n-1之间的，比较这两个大小，取大即可。

使用dp解决单个问题，第m个房子的最大值取决于第m-1房的最大值和m-2房最大值+m房的值。max(r(m-1), r(m-2)+m)；这里仅使用了两个变量所以不需要使用数组去存储值。

class Solution {public:    int rob(vector<int>& nums) {        int len = nums.size();        if( len<2 ){            return len?nums[0]:0;        }                return max(robber(nums,0,len-2), robber(nums,1,len-1));    }    int robber(vector<int>& nums, int l, int r){        int pre = 0, cur = 0;        for(int i=l; i<=r; i++){            int tmp = max(pre+nums[i], cur);            pre = cur;            cur = tmp;        }        return cur;    }};