How to implement #pop for a Linked List in Ruby

I'm struggling to find a solution for implementing #pop on a LinkedList structure in Ruby. I've currently got my implementation to the point of removing the last element (by setting the @next_node element to nil on the penultimate node) and returning the new last element, but I need my method to remember and return all of the previous nodes that I've looped through also, and I'm struggling to find a way to do this. Can anybody please help me:

LinkedList example structure;

"#<LinkedList:0x000000010ced2508 @head=#<Node:0x000000010ced22b0 @value=13, @next_node=#<Node:0x000000010ced22d8 @value=2, @next_node=#<Node:0x000000010ced2300 @value=25, @next_node=#<Node:0x000000010ced2328 @value=20, @next_node=nil>>>>>"

My current #pop implementation:

  def pop
    return if head.nil?
    current_node = head
    current_node = current_node.next_node until current_node.next_node.next_node.nil?
    current_node.next_node = nil
    return current_node
  end

Updated working method following @tom-lord's answer:

def pop
    if head.nil?
      return head
    elsif head.next_node.nil?
      self.head = nil
      return head
    else
      second_last_node = head
      second_last_node = second_last_node.next_node until second_last_node.next_node.next_node.nil?
  
      last_node = second_last_node.next_node
      second_last_node.next_node = nil
      return last_node
    end
  end

Here is your code updated with some comments to recap what you've done:

def pop
  # Handles the edge case where the list is empty -- fine.
  return if head.nil?

  # Sets `current_node` to the **SECOND-LAST** node.
  current_node = head
  current_node = current_node.next_node until current_node.next_node.next_node.nil?

  # Deletes the SECOND-LAST node's pointer
  current_node.next_node = nil
  
  # Returns the SECOND-LAST node (???!!)
  return current_node
end

There are two things going wrong here:

1. You are returning the second-last node, not the last node.
2. Your code will fail for a linked list that contains exactly 1 node.

I'll focus on fixing (1), and leave (2) as an extension for you to figure out.

def pop
  # Handles the edge case where the list is empty -- fine.
  return if head.nil?

  # Using a more descriptive variable name makes it clearer what's happening
  second_last_node = head
  second_last_node = second_last_node.next_node until second_last_node.next_node.next_node.nil?

  last_node = second_last_node.next_node
  second_last_node.next_node = nil
  
  return last_node
end