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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/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
    int getLength(ListNode *head) {
        int ans = 0;
        for (; head; head = head->next) ++ans;
        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/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
    ListNode *reverseList(ListNode *head) {
        ListNode dummy;
        while (head) {
            auto p = head;
            head = head->next;
            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;
    }
};