Check if two linked lists are equal in Ruby? - ruby

Question

I have the following implementation of a linked list in Ruby:
class Node
attr_accessor :data, :next
def initialize(data = nil)
#data = data
#next = nil
end
end
class LinkedList
def initialize(items)
#head = Node.new(items.shift)
items.inject(#head) { |last, data| #tail = last.next = Node.new(data) }
end
def iterate
return nil if #head.nil?
entry = #head
until entry.nil?
yield entry
entry = entry.next
end
end
def equal?(other_list)
#How do I check if all the data for all the elements in one list are the same in the other one?
end
end
I have tried using the .iterate like this:
def equals?(other_list)
other_list.iterate do |ol|
self.iterate do |sl|
if ol.data != sl.data
return false
end
end
end
return true
end
But this is doing a nested approach. I fail to see how to do it.

Answer 1

You can't do it easily with the methods you have defined currently, as there is no way to access a single next element. Also, it would be extremely useful if you implemented each instead of iterate, which then gives you the whole power of the Enumerable mixin.
class LinkedList
include Enumerable # THIS allows you to use `zip` :)
class Node # THIS because you didn't give us your Node
attr_accessor :next, :value
def initialize(value)
#value = value
#next = nil
end
end
def initialize(items)
#head = Node.new(items.shift)
items.inject(#head) { |last, data| #tail = last.next = Node.new(data) }
end
def each
return enum_for(__method__) unless block_given? # THIS allows block or blockless calls
return if #head.nil?
entry = #head
until entry.nil?
yield entry.value # THIS yields node values instead of nodes
entry = entry.next
end
end
def ==(other_list)
# and finally THIS - get pairs from self and other, and make sure all are equal
zip(other_list).all? { |a, b| a == b }
end
end
a = LinkedList.new([1, 2, 3])
b = LinkedList.new([1, 2, 3])
c = LinkedList.new([1, 2])
puts a == b # => true
puts a == c # => false
EDIT: I missed this on the first run through: equal? is supposed to be referential identity, i.e. two variables are equal? if they contain the reference to the same object. You should not redefine that method, even though it is possible. Rather, == is the general common-language meaning of "equal" as in "having the same value", so I changed it to that.

Answer 2

I think there is something wrong with your initialize method in LinkedList, regardless could this be what you need
...
def equal?(other_list)
other_index = 0
cur_index = 0
hash = Hash.new
other_list.iterate do |ol|
hash[ol.data.data] = other_index
other_index += 1
end
self.iterate do |node|
return false if hash[node.data.data] != cur_index
return false if !hash.has_key?(node.data.data)
cur_index += 1
end
return true
end
...
Assuming this is how you use your code
a = Node.new(1)
b = Node.new(2)
c = Node.new(3)
listA = [a,b,c]
aa = Node.new(1)
bb = Node.new(2)
cc = Node.new(3)
listB = [aa,bb,cc]
linkA = LinkedList.new(listA)
linkB = LinkedList.new(listB)
puts linkA.equal?(linkB)