Ruby Singly Linked List Implementation - ruby

Question

I'd like to write an exist function that returns true if a value exists as a node in a linkedlist and returns false otherwise. So far, I have the following code which always returns true. Any help is appreciated:
class SinglyLinkedList
attr_accessor :head, :tail, :count
def initialize
#head = nil
#tail = nil
#count = 0
end
def insert_front(value)
node = Node.new(value)
if #head.nil?
#head = node
#tail = node
else
node.next = #head
#head = node
end
#count +=1
end
def print_first()
puts head.data
end
def print_last()
puts tail.data
end
def exist(value)
walker = #head
until walker.nil?
if walker.data == value
return true
end
walker = walker.next
end
false
end
def size()
count
end
end
class Node
attr_accessor :data, :next
def initialize(data)
#next = nil
#data = data
end
end
This is my test code:
list = SinglyLinkedList.new
list.insert_front(1)
list.exist(2)
which returns true.

Answer 1

Before solution just consider that singly linked list can insert only via append to list. You try to insert nodes in wrong direction.
The problem with you code is that you use reserved method name next, just refactor you code and write simple:
class SinglyLinkedList
class Node
attr_reader :value
attr_accessor :pointer
def initialize(value, pointer = nil)
#value = value
#pointer = pointer
#count = 0
end
end
attr_reader :head, :tail
def initialize
#head = nil
#tail = nil
end
def insert(value)
node = Node.new(value)
#count += 1
if #head.nil?
#head = node
#tail = node
else
#tail.pointer = node
#tail = node
end
end
def inspect
return [] unless #head
values = []
node = #head
begin
values << node.value
node = node.pointer
end while node != nil
values
end
def exists?(value)
return false unless #head
node = #head
begin
node = node.pointer
end while node != nil && node.value != value
node.nil?
end
end