LeetCode:House Robber

原创已于 2022-07-09 02:51:22 修改 · 838 阅读

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#leetcode #House Robber #dynamic programming

于 2015-11-16 13:41:33 首次发布

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本文探讨了一个经典的动态规划问题——打家劫舍。该问题要求在一个整数数组中找到最大的非相邻元素之和，模拟了抢劫一排房屋但不能连续抢劫相邻两户的情况。文章提供了两种解决方案，一种使用数组进行动态规划，另一种则是经过空间优化后的解决方案。

House Robber

Total Accepted: 41584 Total Submissions: 132317 Difficulty: Easy

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.

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Dynamic Programming

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思路：

1.假设nums=[ 3 2 6 5 6 3]

dp[0]=3,

dp[1]=3;

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

...

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

2.得到 dp=[3 3 9 9 15 15]

code:

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

空间优化后：

class Solution {
public:
    int rob(vector<int>& nums) {
        int len = nums.size();
        if(0==len) return 0;
        int preN=0;
	int preY=0;
        for(int i=1;i<len;i++){
		int tmp = preY;
		preN = max(preN, preY);
		preY = tmp + nums[i];
	}  
        return max(preN, preY);
    }
};