LeetCode213. House Robber II（python）_213. house robber ii python-优快云博客

本文链接：https://blog.youkuaiyun.com/AIpush/article/details/103103003

本文探讨了一种特殊情况下房屋抢劫问题的解决方案，即房屋排列成环形，且相邻房屋有安全系统相连，若同时被抢则会报警。通过将问题分解为两部分并使用动态规划算法，最终找到在不触动警报的情况下，能抢到的最大金额。

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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. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, 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: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

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.

这是一道动态规划算法题，最开始的想法就是dp[i]=dp[i]+max(dp[i-2],dp[i-3])，但是问题在于这是一个首位循环的题，第一个和最后一个不能重复计算，需要考虑在内。

所以我把这个list分成了两部分，一部分是[1:]另一部分是[:-1]也就是一个不算首，一个不算尾，然后再取两者最大值即可。

class Solution:
    def rob(self, nums: List[int]) -> int:

        if len(nums)<3:return max([0]+nums)

        dp=[0]+nums[:-1]
        for i in range(3,len(dp)):
            dp[i]=dp[i]+max(dp[i-2],dp[i-3])

        nums = [0]+nums[1:]
        for i in range(3,len(nums)):
            nums[i]=nums[i]+max(nums[i-2], nums[i-3])
        return max(nums+dp)