Leetcode 918. Maximum Sum Circular Subarray (滑动窗口+单调队列好题)

918. Maximum Sum Circular Subarray
     Medium

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

解法1：
不用滑动窗口和单调队列。实际上我们找连续子数组最大和 与 sum - 连续子数组最小和 之间的最大值就可以了。

class Solution {
public:
    int maxSubarraySumCircular(vector<int>& nums) {
        int sum = 0, currMaxSum = 0, currMinSum = 0;
        int gMaxSum = INT_MIN, gMinSum = INT_MAX;
        for (auto num : nums) {
            sum += num;
            currMaxSum = max(currMaxSum + num, num);
            gMaxSum = max(gMaxSum, currMaxSum);
            currMinSum = min(currMinSum + num, num);
            gMinSum = min(gMinSum, currMinSum);
        }
        //if all negative
        if (gMinSum == sum) return gMaxSum;
        return max(gMaxSum, sum - gMinSum);
    }
};

解法2：滑动窗口 + 单调队列