## Description

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.


## Solutions

抢劫时不能连续抢两家，不然会触发报警。

### 1. DP

在学回溯的时候，看到什么就觉得都该用回溯做，结果纠结半天没想出来。这里只要用动规求解就好了，主要是要分清楚状态转移矩阵：dp[i]=max(nums[i]+dp[i-2], dp[i-1])

要分清求最值的几种方法，DP，BFS，贪心等。

class Solution:
def rob(self, nums: List[int]) -> int:
if not nums or len(nums) <= 0:
return 0
elif len(nums) == 1:
return nums[0]
elif len(nums) == 2:
return max(nums[0], nums[1])
n = len(nums)
dp = [0 for _ in range(n)]
dp[0] = nums[0]
dp[1] = max(nums[0],nums[1])
for i in range(2, len(nums)):
dp[i] = max(dp[i-2] + nums[i], dp[i-1])
return dp[-1]