213. House Robber II

最新推荐文章于 2024-01-18 14:02:27 发布

星夜ya

最新推荐文章于 2024-01-18 14:02:27 发布

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文章标签： Python

本文链接：https://blog.youkuaiyun.com/qq_40102087/article/details/80145491

解决环形排列房屋的最大抢劫数额问题，通过两次迭代避免首尾相连的房屋同时被选中，确保不会触发报警。

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题目：

Note: This is an extension of House Robber.

After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

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.

解析：

这题我一开始是使用C++来做的，所以我就将C++的代码转换为Python了。

这一题其实和House Robber是类似的，只不过这里进行了首尾相连，所以我就计算了两遍，第一次是0到n-2，第二次是1到n-1，因为0和n-1是不能同时取的。

代码：

class Solution:
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        if n == 0:
            return 0
        if n == 1:
            return nums[0]
        pre = 0
        current = 0
        for i in range(0, n-1):
            temp = max(pre + nums[i], current)
            pre = current
            current = temp
        max1 = current
        pre = 0
        current = 0;
        for i in range(1, n):
            temp = max(pre + nums[i], current)
            pre = current
            current = temp
        max2 = current
        return max(max1, max2)