## Description

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),


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.


## Solutions

### 1. DP

# Time: O(n)
# Space: O(n)
class Solution:
def rob(self, nums: List[int]) -> int:
if not nums or len(nums) <= 0:
return 0
elif len(nums) == 1:
return nums
elif len(nums) == 2:
return max(nums, nums)
n = len(nums)
dp0 = [0 for _ in range(n+1)]
dp1 = [0 for _ in range(n+1)]
dp0 = nums
# dp1 = nums
for i in range(2, n+1):
dp0[i] = max(dp0[i-1], dp0[i-2]+nums[i-1])
dp1[i] = max(dp1[i-1], dp1[i-2]+nums[i-1])
return max(dp0[n-1], dp1[n])
# 74/74 cases passed (32 ms)
# Your runtime beats 83.6 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.6 MB)

class Solution:
def rob(self, nums: List[int]) -> int:
if not nums or len(nums) <= 0:
return 0
elif len(nums) == 1:
return nums
elif len(nums) == 2:
return max(nums, nums)
n = len(nums)
dp0 = [0 for _ in range(n)]
dp1 = [0 for _ in range(n)]
dp0 = nums
dp0 = nums
dp1 = nums
for i in range(2, n):
dp0[i] = max(dp0[i-1], dp0[i-2]+nums[i])
dp1[i] = max(dp1[i-1], dp1[i-2]+nums[i])
return max(dp0[-2], dp1[-1])
# 74/74 cases passed (32 ms)
# Your runtime beats 83.6 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)


或者直接在第一题的基础上，传输两个字串，比较大小。