[LeetCode]198.House Robber

题目

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.

思路

这道题的本质相当于在一列数组中取出一个或多个不相邻数，使其和最大。
这是一道动态规划问题。
我们维护一个一位数组dp，其中dp[i]表示到i位置时不相邻数能形成的最大和。
状态转移方程：

dp[0] = num[0] （当i=0时）
dp[1] = max(num[0], num[1]) （当i=1时）
dp[i] = max(num[i] + dp[i - 2], dp[i - 1])   （当i !=0 and i != 1时）

代码

/*-------------------------------------------------------------------
*   日期：2014-04-08
*   作者：SJF0115
*   题目: 198.House Robber
*   来源：https://leetcode.com/problems/house-robber/
*   结果：AC
*   来源：LeetCode
*   总结：
--------------------------------------------------------------------*/
#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int rob(vector<int> &num) {
        if(num.empty()){
            return 0;
        }//if
        int size = num.size();
        vector<int> dp(size,0);
        // dp[i]=max(num[i]+dp[i-2],dp[i-1])
        // dp[i]表示[0,i]取1个或多个不相邻数值的最大收益
        dp[0] = num[0];
        dp[1] = max(num[0],num[1]);
        for(int i = 2;i < size;++i){
            dp[i] = max(dp[i-1],dp[i-2]+num[i]);
        }//for
        return dp[size-1];
    }
};

int main() {
    Solution solution;
    vector<int> num = {4,3,1,3,2};
    cout<<solution.rob(num)<<endl;
    return 0;
}

运行时间