2130.Maximum Twin Sum of a Linked List
===

###### tags: `Medium`,`Linked List`,`Two Pointers`,`Stack`

[2130. Maximum Twin Sum of a Linked List](https://leetcode.com/problems/maximum-twin-sum-of-a-linked-list/)

### 題目描述

In a linked list of size `n`, where `n` is **even**, the i^th^ 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.*

### 範例

**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^

### 解答
#### Python
先貼個基本寫法

class Solution:
    def pairSum(self, head: Optional[ListNode]) -> int:
        arr = []
        
        while head:
            arr.append(head.val)
            head = head.next

        i, j = 0, len(arr) - 1
        mx = 0
        while i < j:
            mx = max(mx, arr[i] + arr[j])
            i += 1
            j -= 1

        return mx

class Solution:
    def pairSum(self, head: Optional[ListNode]) -> int:
        prev, slow, fast = None, head, head

        while fast and fast.next:
            fast = fast.next.next
            temp = slow.next
            slow.next = prev
            prev = slow
            slow = temp

        res = 0
        while slow:
            res = max(res, prev.val + slow.val)
            prev = prev.next
            slow = slow.next
            
        return res

**Fast-slow Pointers**
*Time complexity: $O(N)$
Space complexity: $O(1)$*
> [name=Ron Chen][time=Wed, May 17, 2023]

### Reference

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