918.Maximum Sum Circular Subarray
===

###### tags: `Medium`,`Array`,`DP`,`Divide and Conquer`,`Queue`

[918. Maximum Sum Circular Subarray](https://leetcode.com/problems/maximum-sum-circular-subarray/description/)

### 題目描述

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`.

### 範例

**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 * 10^4^
* -3 * 10^4^ <= `nums[i]` <= 3 * 10^4^

### 解答
#### C++

class Solution {
public:
    int maxSubarraySumCircular(vector<int>& nums) {
        int curMax = INT_MIN, maxSum = 0, curMin = INT_MAX, minSum = 0, total = 0;
        for (auto n : nums) {
            maxSum = max(maxSum, 0) + n;
            curMax = max(curMax, maxSum);
            minSum = min(minSum, 0) + n;
            curMin = min(curMin, minSum);
            total += n;
        }
        return curMax > 0 ? max(curMax, total - curMin) : curMax;
    }
};

> [name=Yen-Chi Chen][time=Thu, Jan 19, 2023]

### Reference

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