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# 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.
```

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
```