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 andit 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 tonightwithout 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.

分析

dp取得是当前最大，所以要延续前面最大的，可能是f[i-2]+nums[i]或者是f[i-1]，哪个大选哪个

class Solution:
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        n = len(nums)
        if n <2:
            return max(nums)

        pp = nums[0]
        p = max(nums[0],nums[1])
        for i in range(2,n):
            cur = max(pp+ nums[i],p)
            p,pp = cur,p

        return p

python可以直接prev和cur一起赋值，所以这里prev是前天，cur是昨天。用前两天的算今天的

class Solution:
    def rob(self, nums: List[int]) -> int:
        prev=cur=0
        for i in nums:
            prev,cur = cur, max(prev+i,cur)
        return cur