leetcode 198. House Robber

最新推荐文章于 2024-01-18 14:02:27 发布

原创最新推荐文章于 2024-01-18 14:02:27 发布 · 323 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

57 篇文章

订阅专栏

本文探讨了一道经典的动态规划问题：一名专业窃贼计划抢劫一条街上的房屋，但为了避免报警，不能连续抢劫相邻的两所房子。文章提供了一个有效的解决方案，通过设定两个数组来记录不同状态下的最大收益。

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.

这种有状态选择的动态规划问题，参考上一题的方法，对每种状态设定一个数组，解法如下：

public class Solution {
    public int rob(int[] nums) {
        if(nums==null||nums.length==0) return 0;
        int[] r = new int[nums.length];
        int[] s = new int[nums.length];

        r[0] = nums[0];
        s[0] = 0;

        for(int i=1; i < nums.length; i++){
            r[i] = Math.max(r[i-1], s[i-1] + nums[i]);
            s[i] = Math.max(r[i-1], s[i-1]);
        }
        return Math.max(r[nums.length-1], s[nums.length-1]);
    }
}