LeetCode in Python 198. House Robber

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)