Day 21 LeetCode 198. House Robber

当想不到今天要做什麽时就来解 LeetCode。

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 a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.nums

Example 1:

Input: nums = [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: nums = [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 3:

Input: nums = [1,2,3]
Output: 3

from typing import List

class Solution:

    def solve(self, nums: List[int]) -> int:

        bag = [0] * (len(nums))

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

    def rob(self, nums: List[int]) -> int:
        if len(nums) <= 2:
            return max(nums)
        return self.solve(nums)
        pass

一样是需要巧思才能快速解的动态规划题目。
这题的重点在於两个袋子，一个是昨天的一个是前天的，
要选择前天的+今天抢的，还是今天跳过这家用昨天的袋子呢...