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.Credits:Special thanks to @Freezen for adding this problem and creating all test cases.

• 这道题目没有太大难度，考虑到首位不能相邻外，基本上按照house robber即可完成。

class Solution {public:    int rob1(vector<int>& nums) {        vector<int> dp(nums.size(),0);        if(nums.size() <= 0){            return 0;        }        dp[0] = nums[0];        for(int i = 1;i < nums.size();++i){            if( i >= 2){               dp[i] = max(dp[i-1],dp[i-2]+nums[i]);             }else{               dp[i] = max(dp[i-1],nums[i]);            }        }        return dp[nums.size()-1];    }    int rob(vector<int>& nums) {        if(nums.size() <= 0){            return 0;        }        if(nums.size() == 1){            return nums[0];        }        if(nums.size() == 2){            return max(nums[0],nums[1]);        }        vector<int> v1;        vector<int> v2;        vector<int> v3;        v1.assign(nums.begin()+2, nums.end()-1);        v2.assign(nums.begin()+1, nums.end()-2);        v3.assign(nums.begin()+1, nums.end()-1);        int max1 = nums[0] + rob1(v1);        int max2 = nums[nums.size()-1] + rob1(v2);        int max3 = rob1(v3);        return max(max1,max(max2,max3));    }};