[leetcode] 198. House Robber @ python

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 · 237 阅读

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本文探讨了一种专业抢劫者如何在不触动报警系统的情况下，从一排房屋中窃取最大金额的问题。通过动态规划的方法，我们定义了状态转移方程，实现了在O(n)的时间复杂度和O(n)的空间复杂度下解决问题。文章提供了详细的算法解释和Python代码实现。

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原题

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.

解法

动态规划. 初始化dp[i]为到nums[i]时最多能抢到的钱, 状态转移方程:

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

Time: O(n)
Space: O(n)

代码

class Solution:
    def rob(self, nums: 'List[int]') -> 'int':
        # base case
        if not nums: return 0
        if len(nums) <= 2: return max(nums)
        
        dp = [0]*len(nums)
        dp[0], dp[1] = nums[0], max(nums[0], nums[1])
        for i in range(2, len(nums)):
            dp[i] = max(dp[i-1], dp[i-2] + nums[i])
        return dp[-1]