> For the complete documentation index, see [llms.txt](https://nataliekung.gitbook.io/ladder_code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nataliekung.gitbook.io/ladder_code/qiang-hua-7-dp/house-robber-ii.md).

# House Robber II

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 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:
 [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:
 [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.
```

分析

拆环：1.去头一个数组，去尾一个数组 2.double数组，然后一个window，长度是数组长度

注意这里yes 和No交换进行的方法

```
class Solution:
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0


        n = len(nums)
        if n == 1:
            return nums[0]
        nums1 = nums[:n-1]
        nums2 = nums[1:]
        f1 = self.canDo(nums1)
        f2 = self.canDo(nums2)
        return max(f1,f2)

    def canDo(self, nums):        
        r = len(nums)
        ppre=pre=cur = 0

        for i in range(r):
            cur = max(pre,ppre+nums[i])
            ppre,pre = pre,cur

        return cur
```