lc198. House Robber

最新推荐文章于 2025-08-20 16:36:45 发布

转载最新推荐文章于 2025-08-20 16:36:45 发布 · 84 阅读

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原文链接：https://juejin.im/post/5ce4ea3df265da1bc94ec481

文章标签：

#数据结构与算法 #python

博客围绕House Robber问题展开，给定代表每间房屋藏钱数的非负整数列表，需确定不触发警报情况下可抢劫的最大金额。采用动态规划思路，用字典存储不同情况返回值，同时要注意边界值，还给出了Python3代码示例。

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House Robber

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.

Example 1:

Input: [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: [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.

思路：动态规划，参考 www.cnblogs.com/grandyang/p… 字典存1，2，3等各种情况返回的值注意边界值

f(1) = num[0]
f(2) = max(num[0],num[1])
f(3) = max(num[0]+num[2],num[1])
f(n) = max(num[n-1]+f(n-2),f(n-1))
复制代码

代码:python3

class Solution:
    def rob(self, nums: List[int]) -> int:
        dp={}
        if len(nums) == 0:
            return 0
        if len(nums) == 1:
            return nums[0]
        if len(nums) == 2:
            return max(nums[0],nums[1])
        dp[0] = nums[0]    
        dp[1] = max(nums[0],nums[1])
        for i in range(2,len(nums)):
            dp[i]=max(dp[i-2]+nums[i],dp[i-1])
        return dp[len(nums)-1]
复制代码

转载于:https://juejin.im/post/5ce4ea3df265da1bc94ec481