# 918. Maximum Sum Circular Subarray

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
``````

Note:

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

``````class Solution {
public:
int maxSubarraySumCircular(vector<int>& A) {
int sum = 0, mn = INT_MAX, mx = INT_MIN, curMax = 0, curMin = 0;
for (int num : A) {
curMin = min(curMin + num, num);
mn = min(mn, curMin);
curMax = max(curMax + num, num);
mx = max(mx, curMax);
sum += num;
}
return (sum - mn == 0) ? mx : max(mx, sum - mn);
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/918

Maximum Subarray

Circular Array Loop

https://leetcode.com/problems/maximum-sum-circular-subarray/

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

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