LeetCode198. House Robber_f(k) = max(f(k-1),nums[k-1]+f(k-2))-优快云博客

本文链接：https://blog.youkuaiyun.com/AIpush/article/details/84844522

本文探讨了一种动态规划算法，解决专业窃贼计划抢劫不相邻房屋以最大化收益的问题。通过实例展示了如何避免触发报警系统，同时计算出最大可窃金额。

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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, 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:跟之前上楼梯的问题类似,这是一道动态规划的算法问题.因为不能相连,所以要记录三个数据,1是到上一个数为止的最大值,2是到现在为止的最大值,3是下一个要读取的数字.然后用递归的方式,一直遍历完列表.

f(0) = nums[0]
f(1) = max(num[0], num[1])
f(k) = max( f(k-2) + nums[k], f(k-1) )

其中同时给多个变量赋值可能跟想的不太一样,具体算法可以参考我前一篇博文

class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        last = 0
        now = 0
        
        for num in nums:
            last, now = now, max(last+num,now)

        return now