918. Maximum Sum Circular Subarray

Given a circular integer array nums of length n, return _the maximum possible sum of a non-empty subarray of _nums.

A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].

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

Example 1:

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

Example 2:

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

Example 3:

Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.

Constraints:

• n == nums.length
• 1 <= n <= 3 * 104
• -3 * 104 <= nums[i] <= 3 * 104

class Solution {
public:
    int maxSubarraySumCircular(vector<int>& nums) {
        int curMax = 0, curMin = 0;
        int maxSum = nums[0], minSum = nums[0];
        int totalSum = 0;

        for(int num : nums) {
            curMax = max(curMax + num, num);
            maxSum = max(maxSum, curMax);

            curMin = min(curMin + num, num);
            minSum = min(minSum, curMin);
            
            totalSum += num;
        }
        if(totalSum == minSum) {
            return maxSum;
        }
        return max(maxSum, totalSum - minSum);
    }
};

• 時間複雜度：
  $O(N)$
• 空間複雜度：
  $O(1)$