2130. Maximum Twin Sum of a Linked List.cpp

In a linked list of size n, where n is even, the ith node (0-indexed) of the linked list
is known as the twin of the (n-1-i)th node, if 0 <= i <= (n / 2) - 1.

For example, if n = 4, then node 0 is the twin of node 3, and node 1 is the twin of node 2.
These are the only nodes with twins for n = 4.
The twin sum is defined as the sum of a node and its twin.

Given the head of a linked list with even length, return the maximum twin sum of the linked list.

Example 1:
Input: head = [5,4,2,1]
Output: 6
Explanation:
Nodes 0 and 1 are the twins of nodes 3 and 2, respectively. All have twin sum = 6.
There are no other nodes with twins in the linked list.
Thus, the maximum twin sum of the linked list is 6. 

Example 2:
Input: head = [4,2,2,3]
Output: 7
Explanation:
The nodes with twins present in this linked list are:
- Node 0 is the twin of node 3 having a twin sum of 4 + 3 = 7.
- Node 1 is the twin of node 2 having a twin sum of 2 + 2 = 4.
Thus, the maximum twin sum of the linked list is max(7, 4) = 7.

Example 3:
Input: head = [1,100000]
Output: 100001
Explanation:
There is only one node with a twin in the linked list having twin sum of 1 + 100000 = 100001.

Constraints:
The number of nodes in the list is an even integer in the range [2, 10^5].
1 <= Node.val <= 10^5
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# Approach 1:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* getMiddle(ListNode *head) {
        if (!head) return head;
        
        ListNode *slow = head, *fast = head;

        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }

    int pairSum(ListNode* head) {
        if (!head) return 0;

        stack <int> s;
        int res = INT_MIN;

        ListNode *mid = getMiddle(head);
        ListNode *curr = mid;

        while (curr) {
            s.push(curr->val);
            curr = curr->next;
        }

        curr = head;

        while (!s.empty()) {
            res = max(res, curr->val + s.top());
            s.pop();
            curr = curr->next;
        }

        return res;
    }
};

TC -> O(n), n is the length of the linked list
SC -> O(n), n is the length of the linked list