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

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

Note:

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

思路：要麼沒有橫跨首尾，要麼橫跨；

沒有橫跨就是running sum求最大；橫跨的話就求剩餘部分的最小running sum

class Solution(object):
    def maxSubarraySumCircular(self, a):
        """
        :type A: List[int]
        :rtype: int
        """
        if all(t<0 for t in a): return max(a)
        ma_,mi_=float('-inf'),float('inf')
        ma,mi=float('-inf'),float('inf')
        s=su=0
        for i in a:
            s+=i
            ma_=max(ma_,s)
            mi_=min(mi_,s)
            ma=max(ma,s-mi_)
            mi=min(mi,s-ma_)
            su+=i
        return max(ma,su-mi)
    
s=Solution()
print(s.maxSubarraySumCircular([1,-2,3,-2]))
print(s.maxSubarraySumCircular([5,-3,5]))
print(s.maxSubarraySumCircular([3,-1,2,-1]))
print(s.maxSubarraySumCircular([3,-2,2,-3]))
print(s.maxSubarraySumCircular([-2,-3,-1]))