213. House Robber II -------------M

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

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.

Example 1:

Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

A:
递归，with memory

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> V(n, vector<int>(n,-1));
        return helper(nums,0,n-1, V);
    }
private:
    int helper(vector<int> & nums, int start, int end, vector<vector<int>> & V){
        if(start>end)
            return 0;
        if(V[start][end]>=0)
            return V[start][end];
        int withStart = nums[start] + helper(nums, start+2, end - (start==0?1:0), V);
        int noStart = helper(nums, start+1, end, V);
        
        int res = max(withStart, noStart);
        V[start][end] = res;
        return res;
    }
};

思路很简单，就是每次不取  头  或者不取尾。

class Solution {
public:
    int rob(vector<int>& nums) {        
        int n = nums.size();
        if(n==1)
            return nums[0];
        // two cases, with, or without first element
        return max(helper(nums,0, n-2), helper(nums, 1, n-1) );
    }
private:
    int helper(vector<int> & nums, int start, int end){
        //DP, each time, we jump or not jump
        int cur = 0, pre = 0; // current maxSum, or preMaxSum
        for(int i =start; i<= end; i++){
            int newMax = max(cur, pre+nums[i]);
            pre = cur;
            cur = newMax;
        }
        return cur;
    }
};

Mistakes:

忘记了当只有一个的情况，要单独列出来