Description:

Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C.

Here, a circular array means the end of the array connects to the beginning of the array. (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)

Also, a subarray may only include each element of the fixed buffer A at most once. (Formally, for a subarray C[i], C[i+1], ..., C[j], there does not exist i <= k1, k2 <= j with k1 % A.length = k2 % A.length.)

Example 1:

Input: [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3

Example 2:

Input: [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10

Example 3:

Input: [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4

Example 4:

Input: [3,-2,2,-3]
Output: 3
Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3

Example 5:

Input: [-2,-3,-1]
Output: -1
Explanation: Subarray [-1] has maximum sum -1

Constraints:

1. -30000 <= A[i] <= 30000
2. <= A.length <= 30000

注意这里的数组是连成环的，因此最大子数组存在两种情况：

1. 在数组内；
2. 在数组的边缘部分，即数组的头尾相连
   

对于第2种情况，即将数组的和减去最小和的子数组即可。这里会使用到求解最大子数组和的Kadane算法，不懂得可以先了解以下这个算法。

class Solution:
    def maxSubarraySumCircular(self, A: List[int]) -> int:
        cur_min, cur_max, max_sum, min_sum, total = 0, 0, -pow(10, 10), pow(10, 10), 0
        for num in A:
            cur_max = max(cur_max+num, num)
            max_sum = max(max_sum, cur_max)
            
            cur_min = min(cur_min+num, num)
            min_sum = min(min_sum, cur_min)
            
            total += num
        # 对于max_sum < 0的情况，total-min_sum的结果会为0
        return max(max_sum, total-min_sum) if max_sum > 0 else max_sum

参考资料

1. https://leetcode.com/problems/maximum-sum-circular-subarray/
2. https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/One-Pass