Leetcode 918. Maximum Sum Circular Subarray (单调队列好题)

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：单调队列+滑动窗口。但很不容易调通。
注意这里的滑动窗口的left就是dq.front，右窗口就是i。也就是说，滑动窗口的大小就相当于是i - dq.front。
什么时候调整左窗口? while (i - dq.front() > n) dq.pop_front();文章来源地址https://uudwc.com/A/dbyWg

class Solution {
public:
    /**
     * @param a: the array
     * @return: Maximum Sum Circular Subarray
     */
    int maxSubarraySumCircular(vector<int> &a) {
        int n = a.size(), n2 = (n << 1);
        int res = INT_MIN;
        deque<int> dq; //monotonous queue
        vector<int> nums2(n2, 0);
        vector<int> presums(n2 + 1, 0);
        for (int i = 0; i < n2; i++) {
            nums2[i] = a[i % n];
        }        
        for (int i = 1; i <= n2; i++) {
            presums[i] = presums[i - 1] + nums2[i - 1];
        }
        for (int i = 1; i <= n2; i++) {
            while (!dq.empty() && presums[dq.back()] > presums[i]) {
                dq.pop_back();
            }
            dq.push_back(i);
            while (i - dq.front() > n) dq.pop_front();
            
            if (!dq.empty()) {
                if (i == dq.front()) res = max(res, presums[i] - presums[i - 1]); //当输入全为负数时，处理不一样。
                else res = max(res, presums[i] - presums[dq.front()]);
            } else {
                res = max(res, presums[i]);
            }

        }
        return res;
    }
};