Swapping Nodes in a Linked List

You are given the head of a linked list, and an integer k.

Return the head of the linked list after swapping the values of the kth node from the beginning and the kth node from the end (the list is 1-indexed).

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [1,4,3,2,5]

Example 2:

Input: head = [7,9,6,6,7,8,3,0,9,5], k = 5
Output: [7,9,6,6,8,7,3,0,9,5]

Constraints:

• The number of nodes in the list is n.
• 1 <= k <= n <= 105
• 0 <= Node.val <= 100

SOLUTION:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def lenList(self, head, a, k):
        if head:
            curr = 1 + self.lenList(head.next, a + 1, k)
            if a == k:
                self.a = head
            if curr == k:
                self.b = head
            return curr
        return 0

    def swapNodes(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        n = self.lenList(head, 1, k)
        self.a.val, self.b.val = self.b.val, self.a.val
        return head