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# 題目敘述

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.

# Solution

/**
 * 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:
    int pairSum(ListNode* head) {
        vector<int> left;
        ListNode *slow = head, *fast = head->next;
        left.push_back(slow->val);
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
            left.push_back(slow->val);
        }
        slow = slow->next;
        int res = 0;
        reverse(left.begin(), left.end());
        for (int i=0; i<left.size(); i++) {
            res = max(res, left[i] + slow->val);
            slow = slow->next;
        }
        return res;
    }
};
class Solution {
    public int pairSum(ListNode head) {
        Deque<Integer> dq = new LinkedList<>();
        ListNode curr = new ListNode();
        curr = head;
        while(curr != null){
            dq.add(curr.val);
            curr = curr.next;
        }
        int max = Integer.MIN_VALUE;
        while(!dq.isEmpty() && dq.size() >= 2){
            max = Math.max(max, dq.pollFirst() + dq.pollLast());
        }
        return max;
    }
}