House Robber

1，题目要求

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that 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: [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.

Example 2:
Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

你是一个专业的强盗，计划在街上抢劫房屋。 每个房子都有一定数量的钱存在，阻止你抢劫他们的唯一限制是相邻的房屋有连接的安全系统，如果两个相邻的房子在同一个晚上被打破，它将自动联系警察。

给出一个代表每个房子的金额的非负整数列表，确定今晚可以抢劫的最大金额而不警告警察。

2，题目思路

对于这道题，很明显的，又是一道关于动态规划的题目。

题目的意思稍微有一点复杂，主要就是说对于一个数组，从中取的数字两两不可以相互挨着。如何取数字，使得它们的和最大。

当我们使用动态规划时，首先要对问题进行分析，设置动态规划数dp组用来存储中间结果

• dp[0] = nums[0]
• dp[1] = max( nums[0], nums[1])
• dp[2] = max(dp[0] + nums[2], dp[1])

其形式经过分析之后，非常类似于爬楼梯问题——也是动态规划的较为经典的问题之一。

• dp[i] = max(dp[i-2] + nums[i], dp[i-1])

3，代码实现

int x = []() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    return 0;
}();


class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        
        if(n == 0)
            return 0;
        if(n == 1)
            return nums[0];
        if(n == 2)
            return max(nums[0], nums[1]);
        
        vector<int> dp (n, 0);
        
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);
        for(int i = 2;i<n;i++)
            dp[i] = max(dp[i-2] + nums[i], dp[i-1]);
        
        
        return dp[n-1];
        }
};