LeetCode in Python 198. House Robber

最新推荐文章于 2020-05-29 09:25:03 发布

weixin_30662011

最新推荐文章于 2020-05-29 09:25:03 发布

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文章标签： python

原文链接：http://www.cnblogs.com/lowkeysingsing/p/11274868.html

本文探讨了一种经典的动态规划问题，即如何在一个有安全系统的街区中，作为专业窃贼，选择最佳的房屋进行抢劫，以达到最大的收益，同时避免因连续抢劫相邻房屋而触发警报。文章通过具体示例，详细解释了算法的实现过程。

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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.

不可偷取两个相邻的，求最大和。dp即可，dp[i]表示以i为终点的最大和，dp[i]=nums[i]+max(dp[i-2],dp[i-3])，需要在0之前补位3个，所以nums[i]对应nums[i-3]

class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        
        if not nums: return 0
        
        size = len(nums) + 3
        dp = [0] * size
        for i in range(3, len(dp)):
            dp[i] = nums[i-3] + max(dp[i-2], dp[i-3])
            
        return max(dp)

转载于:https://www.cnblogs.com/lowkeysingsing/p/11274868.html