leetcode 198. House Robber 入室抢劫 + DP动态规划求解-蒲公英云

leetcode 198. House Robber 入室抢劫 + DP动态规划求解

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.

题意这样的，抢到不可以连续抢劫，也就是抢劫了 i 那么就意味着 i-1 和 i+1 不可以抢劫，否者警铃就会响起。

这是一个很简单的DP做法，很值得学习。

本题是上一道题leetcode 740. Delete and Earn 动态规划DP、leetcode 213. House Robber II 入室抢劫抢劫问题 + 一道经典的DP动态规划问题和 leetcode 337. House Robber III DP动态规划 + DFS深度有限遍历

代码如下：

/*
 * https://segmentfault.com/a/1190000005138379
 * 
 * 这个问题很简单是一个简单的DP算法，张辉你对DP还需要多多学习
 * 
 * */
public class Solution 
{
    public int rob(int[] nums) 
    {
        if(nums==null || nums.length<=0)
            return 0;
        if(nums.length==1)
            return nums[0];
        else if(nums.length==2)
            return Math.max(nums[0], nums[1]);
        else
        {
            //dp[i]表示抢劫到i家的最多的钱数
            int []dp=new int[nums.length];
            dp[0]=nums[0];
            dp[1]=Math.max(dp[0], nums[1]);
            for(int i=2;i<nums.length;i++)
                dp[i] = Math.max(dp[i-1], dp[i-2]+nums[i]);
            return dp[nums.length-1];
        }
    }
}

下面是C++的做法，这是最经典的DP动态规划的做法

代码如下：

#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <cmath>
#include <algorithm>
using namespace std;
class Solution 
{
public:
    int rob(vector<int>& a) 
    {
        if (a.size() <= 0)
            return 0;
        vector<int> dp(a.size(),0);
        dp[0] = a[0];
        dp[1] = max(a[0], a[1]);
        for (int i = 2; i < a.size(); i++)
        {
            dp[i] = max(a[i] + dp[i - 2], dp[i - 1]);
        }
        return dp[a.size() - 1];
    }
};