213. House Robber II

题意：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.

思路：这题是198. House Robber的扩展，把之前的街区从线段变成了环，具体思路还是没变，因为第一家和最后一家能且必须选择一家，所以把这个环切分成两个线段，分别包含第一家和最后一家，代码：

class Solution {public:    int rob(vector<int>& nums) {        if (nums.size() == 0)            return 0;        else if (nums.size() == 1)            return nums[0];        int last1 = 0, now1 = 0, tmp;        int m = nums.size();        for (int i = 1;i < m;i++){            tmp = now1;            now1 = max(last1+nums[i],now1);            last1 = tmp;        }        int last2 = 0, now2 = 0;        for (int i = 0;i < m - 1;i++){            tmp = now2;            now2 = max(last2+nums[i],now2);            last2 = tmp;        }        return max(now1,now2);    }};