[leetcode] 198. House Robber
2022/3/9 23:19:37
本文主要是介绍[leetcode] 198. House Robber,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
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 systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given an integer array nums
representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: nums = [1,2,3,1] Output: 4 Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.
Example 2:
Input: nums = [2,7,9,3,1] Output: 12 Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). Total amount you can rob = 2 + 9 + 1 = 12.
Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 400
思路
动态规划,当前获取最多的钱等于max(前一次获取最多的钱,前前次获取最多的钱+当前房子的钱)。
代码
python版本:
class Solution: def rob(self, nums: List[int]) -> int: dp = [0 for _ in range(len(nums)+1)] dp[1] = nums[0] for i in range(2, len(dp)): dp[i] = max(dp[i-1], dp[i-2]+nums[i-1]) return dp[len(nums)]
这篇关于[leetcode] 198. House Robber的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!
- 2024-09-28pyqt 怎么打包整个项目-icode9专业技术文章分享
- 2024-09-28laravel Commands 创建带有参数的 Artisan 命令的步骤和示例-icode9专业技术文章分享
- 2024-09-28antd怎么实现渲染tiff图片-icode9专业技术文章分享
- 2024-09-28英文半角中划线和中文全角的中划线有什么区别-icode9专业技术文章分享
- 2024-09-28nvm npm 和node 他们之间有什么关系-icode9专业技术文章分享
- 2024-09-28Node Version Manager (nvm)使用教程-icode9专业技术文章分享
- 2024-09-28nvm命令太慢,是什么原因-icode9专业技术文章分享
- 2024-09-28Kotlin 如何增加、删除和修改 MutableStateFlow 中的值。-icode9专业技术文章分享
- 2024-09-28Kotlin的stateFlow.update 写法介绍-icode9专业技术文章分享
- 2024-09-28kotlin 怎么获取当前时间格式-icode9专业技术文章分享