LeetCode in Elixir
206. Reverse Linked List
Easy
Given the head of a singly linked list, reverse the list, and return the reversed list.
Example 1:
Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
Example 2:
Input: head = [1,2]
Output: [2,1]
Example 3:
Input: head = []
Output: []
Constraints:
- The number of nodes in the list is the range
[0, 5000]. -5000 <= Node.val <= 5000
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
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_list(head :: ListNode.t | nil) :: ListNode.t | nil
def reverse_list(nil), do: nil
def reverse_list(%ListNode{val: v, next: n}), do: reverse_list(n, %ListNode{val: v, next: nil})
def reverse_list(%ListNode{val: v, next: n}, acc), do: reverse_list(n, %ListNode{val: v, next: acc})
def reverse_list(nil, acc), do: acc
end