leetcode 198 House Robber

jzman 发布于2019-08-16 11:06 / 769人阅读

摘要：但是任何临近的两个房子被偷就会触发警报。要求我们求出在不触发警报的情况下偷到的最多的钱。每个房子里的钱通过输入的数组表示。

题目详情

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.

题目的意思是，我们是一个江洋大盗～现在我们要去偷整条街的房子，每个房子里有一定的钱。但是任何临近的两个房子被偷就会触发警报。要求我们求出在不触发警报的情况下偷到的最多的钱。每个房子里的钱通过输入的int数组表示。

想法

动态规划问题

对于每一个房子，我们有偷／不偷两种选择

因此我们声明两个变量prevNo和prevYes分别保存，我没偷/偷了当前房子的情况下，目前为止偷的最多的钱数。

如果想偷当前房子，那么要求我们并没有偷前一个房子，所以用前一个房子的prevNo值和当前房子的钱数相加。

如果不偷当前房子，那我们可以取前一个房子的prevNo值和prevYes值中较大的那个。

解法

    public int rob(int[] nums) {
                int prevNo = 0;
        int prevYes = 0;
        
        for(int n: nums){
            int temp = prevNo;
            prevNo = Math.max(prevNo, prevYes);
            prevYes = temp+n;
        }
        
        return Math.max(prevNo, prevYes);
    }

云服务器 GPU云服务器 Robber House 198cdn Leetcod

文章版权归作者所有，未经允许请勿转载,若此文章存在违规行为，您可以联系管理员删除。

转载请注明本文地址：https://www.ucloud.cn/yun/71012.html

leetcode198,213 house robber

摘要：你不能连着偷两家因为这样会触发警报系统。现在有一个数组存放着每一家中的可偷金额，问可以偷的最大金额为多少这里考验了动态编程的思想。动态编程要求我们将问题一般化，然后再找到初始情况开始这个由一般到特殊的计算过程。 House Robber I You are a professional robber planning to rob houses along a street. Each...

whidy 2019-08-16 11:01 评论0 收藏0
[LeetCode] House Robber I II

摘要：注意对边界条件的判断，是否非空，是否长度为 House Robber I Problem 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 y...

qpal 2019-08-14 15:17 评论0 收藏0
[Leetcode] House Robber 打家劫舍

摘要：动态规划复杂度时间空间思路一般来说，给定一个规则，让我们求任意状态下的解，都是用动态规划。另外我们可以做一点优化，本来我们是要用一个数组来保存之前的结果的。所以我们分别算出这两个条件下的最大收益，然后取更大的就行了。可以复用的代码。 House Robber I You are a professional robber planning to rob houses along a ...

golden_hamster 2019-08-14 12:38 评论0 收藏0
LeetCode[337] House Robber III

摘要：复杂度思路对于每一个位置来说，考虑两种情况分别对和再进行计算。用对已经计算过的进行保留，避免重复计算。 LeetCode[337] House Robber III The thief has found himself a new place for his thievery again. There is only one entrance to this area, calle...

Dr_Noooo 2019-08-14 15:00 评论0 收藏0
[LintCode/LeetCode] House Robber II

摘要：因为取了队首就不能取队尾，所以对数组循环两次，一个从取到，一个从取到，比较两次循环后队尾元素，取较大者。注意，要先讨论当原数组位数小于时的情况。 Problem After robbing those houses on that street, the thief has found himself a new place for his thievery so that he wi...

OnlyLing 2019-08-14 15:34 评论0 收藏0