Formatted question description: https://leetcode.ca/all/2130.html

# 2130. Maximum Twin Sum of a Linked List (Medium)

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, 105].
• 1 <= Node.val <= 105

Related Topics:
Linked List, Two Pointers, Stack

Similar Questions:

## Solution 1.

// OJ: https://leetcode.com/problems/maximum-twin-sum-of-a-linked-list/
// Time: O(N)
// Space: O(N)
class Solution {
int getLength(ListNode *head) {
int ans = 0;
return ans;
}
public:
int pairSum(ListNode* head) {
int len = getLength(head), ans = 0;
vector<int> sum(len / 2);
for (int i = 0; head; head = head->next, ++i) {
ans = max(ans, sum[i < len / 2 ? i : (len - 1 - i)] += head->val);
}
return ans;
}
};


## Solution 2.

// OJ: https://leetcode.com/problems/maximum-twin-sum-of-a-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
ListNode *reverseList(ListNode *head) {
ListNode dummy;
auto p = head;
p->next = dummy.next;
dummy.next = p;
}
return dummy.next;
}
public:
int pairSum(ListNode* head) {
auto p = head, q = head;
while (q->next && q->next->next) {
p = p->next;
q = q->next->next;
}
q = p->next;
p->next = nullptr;
q = reverseList(q);
int ans = 0;
for (; head; head = head->next, q = q->next) {
ans = max(ans, head->val + q->val);
}
return ans;
}
};