Brandon Linked_List Solution

linked_list.rb

class LinkedListNode
  attr_accessor :value, :next_node

  def initialize(value, next_node=nil)
    @value = value
    @next_node = next_node
  end
end

def print_values(list_node)
  print "#{list_node.value} --> "
  if list_node.next_node.nil?
    print "nil\n"
    return
  else
    print_values(list_node.next_node)
  end
end
        # list = node3 / node2 / node1
def reverse_list(list, previous=nil)     # / <-represents each recursion when reverse_list is called inside method
  new_head = list.next_node      # new_head = node2           / new_head = node1           / new_head = nil
  list.next_node = previous      # node3.next_node = nil      / node2.next_node = node3.   / node1.next_node = node2
  if new_head # != nil 
    reverse_list(new_head, list) # reverse_list(node2, node3) / reverse_list(node1, node2) / return list if new_head is nil
  else                                                                                     #         ^ = node1 
    list
  end
end

def check_if_loop(node)
  t = node
  h = node
  while h != nil
    h = h.next_node
    return false if h == nil
    h = h.next_node
    t = t.next_node
    return true if h == t
  end
end
      

node1 = LinkedListNode.new(37)         # node2
node2 = LinkedListNode.new(99, node1)  # node3
node3 = LinkedListNode.new(12, node2)  # nil

# node1.next_node = node3        # <- Makes infinite loop

# puts check_if_loop(node3)      # <- return true or false for infinite loop

print_values(node3)

puts "-------"

revlist = reverse_list(node3)

print_values(revlist)