README.md

1721. 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]

Example 3:

Input: head = [1], k = 1
Output: [1]

Example 4:

Input: head = [1,2], k = 1
Output: [2,1]

Example 5:

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

Constraints:

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

Solutions (Ruby)

1. Solution

# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode} head
# @param {Integer} k
# @return {ListNode}
def swap_nodes(head, k)
  dummy = ListNode.new(0, head)
  curr = head
  n = 0
  until curr.nil?
    n += 1
    curr = curr.next
  end

  curr = dummy
  node0 = nil
  node1 = nil
  (0..n).each do |i|
    node0 = curr if i == k - 1
    node1 = curr if i == n - k
    curr = curr.next
  end

  temp = node0.next
  node0.next = node1.next
  node1.next = temp
  temp = node0.next.next
  node0.next.next = node1.next.next
  node1.next.next = temp

  dummy.next
end