# Doubly Linked List in Python

Here is round two for linked lists, the doubly linked list. It’s not very much more complex than a singly linked list. Breaking things into cases using the `self.count` attribute makes the code easier to read and reason about.

```class DoublyLinkedList:

class Node:
def __init__( self, data, prevNode, nextNode ):
self.data = data
self.prevNode = prevNode
self.nextNode = nextNode

def __init__( self, data=None ):
self.first = None
self.last = None
self.count = 0
if self.count == 0:
self.first = self.Node( data, None, None )
self.last = self.first
elif self.count > 0:
# create a new node pointing to self.first
node = self.Node( data, None, self.first )
# have self.first point back to the new node
self.first.prevNode = node
# finally point to the new node as the self.first
self.first = node
self.current = self.first
self.count += 1
def popFirst( self ):
if self.count == 0:
raise RuntimeError("Cannot pop from an empty linked list")
result = self.first.data
if self.count == 1:
self.first = None
self.last = None
else:
self.first = self.first.nextNode
self.first.prevNode = None
self.current = self.first
self.count -= 1
return result
def popLast( self ):
if self.count == 0:
raise RuntimeError("Cannot pop from an empty linked list")
result = self.last.data
if self.count == 1:
self.first = None
self.last = None
else:
self.last = self.last.prevNode
self.last.nextNode = None
self.count -= 1
return result
if self.count == 0:
else:
self.last.nextNode = self.Node( data, self.last, None )
self.last = self.last.nextNode
self.count += 1
def __repr__( self ):
result = ""
if self.count == 0:
return "..."
cursor = self.first
for i in range( self.count ):
result += "{}".format(cursor.data)
cursor = cursor.nextNode
if cursor is not None:
result += " --- "
return result
def __iter__( self ):
# lets Python know this class is iterable
return self
def next( self ):
# provides things iterating over class with next element
if self.current is None:
# allows us to re-iterate
self.current = self.first
raise StopIteration
else:
result = self.current.data
self.current = self.current.nextNode
return result
def __len__( self ):
return self.count