LeetCode in Elixir

25. Reverse Nodes in k-Group

Hard

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list’s nodes, only nodes themselves may be changed.

Example 1:

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

Output: [2,1,4,3,5]

Example 2:

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

Output: [3,2,1,4,5]

Example 3:

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

Output: [1,2,3,4,5]

Example 4:

Input: head = [1], k = 1

Output: [1]

Constraints:

• The number of nodes in the list is in the range sz.
• 1 <= sz <= 5000
• 0 <= Node.val <= 1000
• 1 <= k <= sz

Follow-up: Can you solve the problem in O(1) extra memory space?

Solution

# Definition for singly-linked list.
#
# defmodule ListNode do
#   @type t :: %__MODULE__{
#           val: integer,
#           next: ListNode.t() | nil
#         }
#   defstruct val: 0, next: nil
# end

defmodule Solution do
  @spec reverse_k_group(head :: ListNode.t() | nil, k :: integer) :: ListNode.t() | nil
  def reverse_k_group(head, k) do
    cond do
      # if length is less then k then do not change it
      length(0, head) < k -> head
      true -> head |> drop(k) |> reverse_k_group(k) |> reverse_list(head, k)
    end
  end

  def reverse_list(acc, _, 0), do: acc

  def reverse_list(acc, list, k) do
    reverse_list(%ListNode{list | next: acc}, list.next, k - 1)
  end

  def drop(nil, _), do: nil
  def drop(head, 0), do: head
  def drop(head, n), do: drop(head.next, n - 1)

  def length(acc, nil), do: acc
  def length(acc, head), do: length(acc + 1, head.next)

end

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