LeetCode OJ：House Robber II（房屋窃贼II）

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 tonight without alerting the police.

这一题和前面的那个窃贼问题比较类似，不过这里首位同样判定是相邻的，所以这里的做法应该是先求出第一个到倒数第二个的最大值，在求出第二个到最后一个的最大值，两者较大的就是最大的值了，同样用dp来解决，代码如下所示：

class Solution {
public:
    int rob(vector<int>& nums) {
        if(!nums.size()) return ;
        if(nums.size() == ) return nums[];
        vector<int> ret1, ret2;
        ret1.resize(nums.size());
        ret2.resize(nums.size());
        ret1[] = nums[];
        ret2[] = nums[];
        for(int i = ; i < nums.size() - ; ++i){
            ret1[i] = max((i ==  ?  : ret1[i - ]) + nums[i], ret1[i - ]);
        }
        for(int i = ; i < nums.size(); ++i){
            ret2[i] = max((i ==  ?  : ret2[i - ]) + nums[i], ret2[i - ]);
        }
        return max(ret1[nums.size() - ], ret2[nums.size() - ]);
    }
};