# How can I use a Linked List in Python?

## Question:

What’s the easiest way to use a linked list in Python? In Scheme, a linked list is defined simply by '(1 2 3 4 5).

Python’s lists, [1, 2, 3, 4, 5], and tuples, (1, 2, 3, 4, 5), are not, in fact, linked lists, and linked lists have some nice properties such as constant-time concatenation, and being able to reference separate parts of them. Make them immutable and they are really easy to work with!

Immutable lists are best represented through two-tuples, with None representing NIL. To allow simple formulation of such lists, you can use this function:

def mklist(*args):
result = None
for element in reversed(args):
result = (element, result)
return result

To work with such lists, I’d rather provide the whole collection of LISP functions (i.e. first, second, nth, etc), than introducing methods.

I wrote this up the other day

#! /usr/bin/env python

class Node(object):
def __init__(self):
self.data = None # contains the data
self.next = None # contains the reference to the next node

def __init__(self):
self.cur_node = None

new_node = Node() # create a new node
new_node.data = data
new_node.next = self.cur_node # link the new node to the 'previous' node.
self.cur_node = new_node #  set the current node to the new one.

def list_print(self):
node = self.cur_node # cant point to ll!
while node:
print node.data
node = node.next

ll.list_print()

When using immutable linked lists, consider using Python’s tuple directly.

ls = (1, 2, 3, 4, 5)

def first(ls): return ls[0]
def rest(ls): return ls[1:]

Its really that ease, and you get to keep the additional funcitons like len(ls), x in ls, etc.

Here’s a slightly more complex version of a linked list class, with a similar interface to python’s sequence types (ie. supports indexing, slicing, concatenation with arbitrary sequences etc). It should have O(1) prepend, doesn’t copy data unless it needs to and can be used pretty interchangably with tuples.

It won’t be as space or time efficient as lisp cons cells, as python classes are obviously a bit more heavyweight (You could improve things slightly with “__slots__ = '_head','_tail'” to reduce memory usage). It will have the desired big O performance characteristics however.

Example of usage:

>>> l = LinkedList([1,2,3,4])
>>> l
LinkedList([1, 2, 3, 4])
(1, LinkedList([2, 3, 4]))

# Prepending is O(1) and can be done with:
LinkedList([0, 1, 2, 3, 4])
# Or prepending arbitrary sequences (Still no copy of l performed):
[-1,0] + l
LinkedList([-1, 0, 1, 2, 3, 4])

# Normal list indexing and slice operations can be performed.
# Again, no copy is made unless needed.
>>> l[1], l[-1], l[2:]
(2, 4, LinkedList([3, 4]))
>>> assert l[2:] is l.next.next

# For cases where the slice stops before the end, or uses a
# non-contiguous range, we do need to create a copy.  However
# this should be transparent to the user.
LinkedList([90, 92, 94, 96, 98])

Implementation:

import itertools

"""Immutable linked list class."""

def __new__(cls, l=[]):
if isinstance(l, LinkedList): return l # Immutable, so no copy needed.
i = iter(l)
try:
except StopIteration:
return cls.EmptyList   # Return empty list singleton.

obj = super(LinkedList, cls).__new__(cls)
obj._tail = tail
return obj

@classmethod
def cons(cls, head, tail):
if not isinstance(tail, cls):
tail = cls(tail)
ll._tail = tail
return ll

# head and tail are not modifiable
@property

@property
def tail(self): return self._tail

def __nonzero__(self): return True

def __len__(self):
return sum(1 for _ in self)

if not self: return other   # () + l = l
start=l = LinkedList(iter(self))  # Create copy, as we'll mutate

while l:
if not l._tail: # Last element?
l._tail = other
break
l = l._tail
return start

return LinkedList(other) + self

def __iter__(self):
x=self
while x:
x=x.tail

def __getitem__(self, idx):
"""Get item at specified index"""
if isinstance(idx, slice):
# Special case: Avoid constructing a new list, or performing O(n) length
# calculation for slices like l[3:].  Since we're immutable, just return
# the appropriate node. This becomes O(start) rather than O(n).
# We can't do this for  more complicated slices however (eg [l:4]
start = idx.start or 0
if (start >= 0) and (idx.stop is None) and (idx.step is None or idx.step == 1):
no_copy_needed=True
else:
length = len(self)  # Need to calc length.
start, stop, step = idx.indices(length)
no_copy_needed = (stop == length) and (step == 1)

if no_copy_needed:
l = self
for i in range(start):
if not l: break # End of list.
l=l.tail
return l
else:
# We need to construct a new list.
if step < 1:  # Need to instantiate list to deal with -ve step
else:
return LinkedList(itertools.islice(iter(self), start, stop, step))
else:
# Non-slice index.
if idx < 0: idx = len(self)+idx
if not self: raise IndexError("list index out of range")
if idx == 0: return self.head
return self.tail[idx-1]

def __mul__(self, n):
if n <= 0: return Nil
l=self
for i in range(n-1): l += self
return l
def __rmul__(self, n): return self * n

# Ideally we should compute the has ourselves rather than construct
# a temporary tuple as below.  I haven't impemented this here
def __hash__(self): return hash(tuple(self))

def __eq__(self, other): return self._cmp(other) == 0
def __ne__(self, other): return not self == other
def __lt__(self, other): return self._cmp(other) < 0
def __gt__(self, other): return self._cmp(other) > 0
def __le__(self, other): return self._cmp(other) <= 0
def __ge__(self, other): return self._cmp(other) >= 0

def _cmp(self, other):
"""Acts as cmp(): -1 for self<other, 0 for equal, 1 for greater"""
if not isinstance(other, LinkedList):
return cmp(LinkedList,type(other))  # Arbitrary ordering.

A, B = iter(self), iter(other)
for a,b in itertools.izip(A,B):
if a<b: return -1
elif a > b: return 1

try:
A.next()
return 1  # a has more items.
except StopIteration: pass

try:
B.next()
return -1  # b has more items.
except StopIteration: pass

return 0  # Lists are equal

def __repr__(self):
return "LinkedList([%s])" % ', '.join(map(repr,self))

"""A singleton representing an empty list."""
def __new__(cls):
return object.__new__(cls)

def __iter__(self): return iter([])
def __nonzero__(self): return False

@property
def head(self): raise IndexError("End of list")

@property
def tail(self): raise IndexError("End of list")

# Create EmptyList singleton
del EmptyList

For some needs, a deque may also be useful. You can add and remove items on both ends of a deque at O(1) cost.

from collections import deque
d = deque([1,2,3,4])

print d
for x in d:
print x
print d.pop(), d

Here is some list functions based on Martin v. Löwis’s representation:

cons   = lambda el, lst: (el, lst)
mklist = lambda *args: reduce(lambda lst, el: cons(el, lst), reversed(args), None)
car = lambda lst: lst[0] if lst else lst
cdr = lambda lst: lst[1] if lst else lst
nth = lambda n, lst: nth(n-1, cdr(lst)) if n > 0 else car(lst)
length  = lambda lst, count=0: length(cdr(lst), count+1) if lst else count
begin   = lambda *args: args[-1]
display = lambda lst: begin(w("%s " % car(lst)), display(cdr(lst))) if lst else w("niln")

where w = sys.stdout.write

Although doubly linked lists are famously used in Raymond Hettinger’s ordered set recipe, singly linked lists have no practical value in Python.

I’ve never used a singly linked list in Python for any problem except educational.

Thomas Watnedal suggested a good educational resource How to Think Like a Computer Scientist, Chapter 17: Linked lists:

A linked list is either:

• the empty list, represented by None, or
• a node that contains a cargo object and a reference to a linked list.

class Node:
def __init__(self, cargo=None, next=None):
self.car = cargo
self.cdr = next
def __str__(self):
return str(self.car)

def display(lst):
if lst:
w("%s " % lst)
display(lst.cdr)
else:
w("niln")

The accepted answer is rather complicated. Here is a more standard design:

L.insert(1)
L.insert(1)
L.insert(2)
L.insert(4)
print L
L.clear()
print L

It is a simple LinkedList class based on the straightforward C++ design and Chapter 17: Linked lists, as recommended by Thomas Watnedal.

class Node:
def __init__(self, value = None, next = None):
self.value = value
self.next = next

def __str__(self):
return 'Node ['+str(self.value)+']'

def __init__(self):
self.first = None
self.last = None

def insert(self, x):
if self.first == None:
self.first = Node(x, None)
self.last = self.first
elif self.last == self.first:
self.last = Node(x, None)
self.first.next = self.last
else:
current = Node(x, None)
self.last.next = current
self.last = current

def __str__(self):
if self.first != None:
current = self.first
out = 'LinkedList [n' +str(current.value) +'n'
while current.next != None:
current = current.next
out += str(current.value) + 'n'
return out + ']'

def clear(self):
self.__init__()

I based this additional function on Nick Stinemates

new_node = Node()
node = self.curr_node
while node:
if node.next == None:
node.next = new_node
new_node.next = None
new_node.data = data
node = node.next

The method he has adds the new node at the beginning while I have seen a lot of implementations which usually add a new node at the end but whatever, it is fun to do.

I think the implementation below fill the bill quite gracefully.

'''singly linked lists, by Yingjie Lan, December 1st, 2011'''

'''Singly linked list, with pythonic features.
The list has pointers to both the first and the last node.'''
__slots__ = ['data', 'next'] #memory efficient
def __init__(self, iterable=(), data=None, next=None):
'''Provide an iterable to make a singly linked list.
Set iterable to None to make a data node for internal use.'''
if iterable is not None:
self.data, self.next = self, None
self.extend(iterable)
else: #a common node
self.data, self.next = data, next

def empty(self):
'''test if the list is empty'''
return self.next is None

def append(self, data):
'''append to the end of list.'''
last = self.data
self.data = last.next = linkst(None, data)
#self.data = last.next

def insert(self, data, index=0):
'''insert data before index.
Raise IndexError if index is out of range'''
curr, cat = self, 0
while cat < index and curr:
curr, cat = curr.next, cat+1
if index<0 or not curr:
raise IndexError(index)
new = linkst(None, data, curr.next)
if curr.next is None: self.data = new
curr.next = new

def reverse(self):
'''reverse the order of list in place'''
current, prev = self.next, None
while current: #what if list is empty?
next = current.next
current.next = prev
prev, current = current, next
if self.next: self.data = self.next
self.next = prev

def delete(self, index=0):
'''remvoe the item at index from the list'''
curr, cat = self, 0
while cat < index and curr.next:
curr, cat = curr.next, cat+1
if index<0 or not curr.next:
raise IndexError(index)
curr.next = curr.next.next
if curr.next is None: #tail
self.data = curr #current == self?

def remove(self, data):
'''remove first occurrence of data.
Raises ValueError if the data is not present.'''
current = self
while current.next: #node to be examined
if data == current.next.data: break
current = current.next #move on
else: raise ValueError(data)
current.next = current.next.next
if current.next is None: #tail
self.data = current #current == self?

def __contains__(self, data):
'''membership test using keyword 'in'.'''
current = self.next
while current:
if data == current.data:
return True
current = current.next
return False

def __iter__(self):
'''iterate through list by for-statements.
return an iterator that must define the __next__ method.'''
itr.next = self.next
return itr #invariance: itr.data == itr

def __next__(self):
'''the for-statement depends on this method
to provide items one by one in the list.
return the next data, and move on.'''
#the invariance is checked so that a linked list
#will not be mistakenly iterated over
if self.data is not self or self.next is None:
raise StopIteration()
next = self.next
self.next = next.next
return next.data

def __repr__(self):
'''string representation of the list'''

def __str__(self):
'''converting the list to a string'''
return '->'.join(str(i) for i in self)

#note: this is NOT the class lab! see file linked.py.
def extend(self, iterable):
'''takes an iterable, and append all items in the iterable
to the end of the list self.'''
last = self.data
for i in iterable:
last.next = linkst(None, i)
last = last.next
self.data = last

def index(self, data):
'''TODO: return first index of data in the list self.
Raises ValueError if the value is not present.'''
#must not convert self to a tuple or any other containers
current, idx = self.next, 0
while current:
if current.data == data: return idx
current, idx = current.next, idx+1
raise ValueError(data)

The following is what I came up with. It’s similer to Riccardo C.’s, in this thread, except it prints the numbers in order instead of in reverse. I also made the LinkedList object a Python Iterator in order to print the list out like you would a normal Python list.

class Node:

def __init__(self, data=None):
self.data = data
self.next = None

def __str__(self):
return str(self.data)

def __init__(self):
self.curr = None
self.tail = None

def __iter__(self):
return self

def next(self):
if self.head and not self.curr:
return self.curr
elif self.curr.next:
self.curr = self.curr.next
return self.curr
else:
raise StopIteration

def append(self, data):
n = Node(data)
self.tail = n
else:
self.tail.next = n
self.tail = self.tail.next

# Add 5 nodes
for i in range(1, 6):
ll.append(i)

# print out the list
for n in ll:
print n

"""
Example output:
1
2
3
4
5
"""
def __init__(self, value):
self.value = value
self.next = None

def insert(self, node):
if not self.next:
self.next = node
else:
self.next.insert(node)

def __str__(self):
if self.next:
return '%s -> %s' % (self.value, str(self.next))
else:
return ' %s ' % self.value

if __name__ == "__main__":
items = ['a', 'b', 'c', 'd', 'e']
ll = None
for item in items:
if ll:
ll.insert(next_ll)
else:
print('[ %s ]' % ll)

I just did this as a fun toy. It should be immutable as long as you don’t touch the underscore-prefixed methods, and it implements a bunch of Python magic like indexing and len.

First of all, I assume you want linked lists. In practice, you can use collections.deque, whose current CPython implementation is a doubly linked list of blocks (each block contains an array of 62 cargo objects). It subsumes linked list’s functionality. You can also search for a C extension called llist on pypi. If you want a pure-Python and easy-to-follow implementation of the linked list ADT, you can take a look at my following minimal implementation.

class Node (object):
""" Node for a linked list. """
def __init__ (self, value, next=None):
self.value = value
self.next = next

""" Linked list ADT implementation using class.
A linked list is a wrapper of a head pointer
that references either None, or a node that contains
a reference to a linked list.
"""
def __init__ (self, iterable=()):
for x in iterable:

def __iter__ (self):
while p is not None:
yield p.value
p = p.next

def prepend (self, x):  # 'appendleft'

def reverse (self):
""" In-place reversal. """
while p is not None:
p0, p = p, p.next

if __name__ == '__main__':
ll.prepend(3); ll.prepend(2)
print list(ll)
ll.reverse()
print list(ll)
class LL(object):
def __init__(self,val):
self.val = val
self.next = None

def pushNodeEnd(self,top,val):
if top is None:
top.val=val
top.next=None
else:
tmp=top
while (tmp.next != None):
tmp=tmp.next
newNode=LL(val)
newNode.next=None
tmp.next=newNode

def pushNodeFront(self,top,val):
if top is None:
top.val=val
top.next=None
else:
newNode=LL(val)
newNode.next=top
top=newNode

def popNodeFront(self,top):
if top is None:
return
else:
sav=top
top=top.next
return sav

def popNodeEnd(self,top):
if top is None:
return
else:
tmp=top
while (tmp.next != None):
prev=tmp
tmp=tmp.next
prev.next=None
return tmp

top=LL(10)
top.pushNodeEnd(top, 20)
top.pushNodeEnd(top, 30)
pop=top.popNodeEnd(top)
print (pop.val)

I’ve put a Python 2.x and 3.x singly-linked list class at https://pypi.python.org/pypi/linked_list_mod/

It’s tested with CPython 2.7, CPython 3.4, Pypy 2.3.1, Pypy3 2.3.1, and Jython 2.7b2, and comes with a nice automated test suite.

It also includes LIFO and FIFO classes.

They aren’t immutable though.

My 2 cents

class Node:
def __init__(self, value=None, next=None):
self.value = value
self.next = next

def __str__(self):
return str(self.value)

def __init__(self):
self.first = None
self.last = None

current = Node(x, None)
try:
self.last.next = current
except AttributeError:
self.first = current
self.last = current
else:
self.last = current

def print_list(self):
node = self.first
while node:
print node.value
node = node.next

ll.print_list()

# Result:
# 1st
# 2nd
# 3rd
# 4th
# 5th
enter code here
enter code here

class node:
def __init__(self):
self.data = None
self.next = None
def __init__(self):
self.cur_node = None
new_node = node()
if self.head == None:
self.cur_node = new_node
new_node.data = data
new_node.next = None
self.cur_node.next = new_node
self.cur_node = new_node
def list_print(self):
while node:
print (node.data)
node = node.next
def delete(self):
next_node = node.next
del(node)
a.delete()
a.list_print()

If you want to just create a simple liked list then refer this code

l=[1,[2,[3,[4,[5,[6,[7,[8,[9,[10]]]]]]]]]]

for visualize execution for this cod Visit http://www.pythontutor.com/visualize.html#mode=edit

'''LIFO Stack implementation using a singly linked list for storage.'''

_ToList = []

#---------- nested _Node class -----------------------------
class _Node:
'''Lightweight, nonpublic class for storing a singly linked node.'''
__slots__ = '_element', '_next'     #streamline memory usage

def __init__(self, element, next):
self._element = element
self._next = next

#--------------- stack methods ---------------------------------
def __init__(self):
'''Create an empty stack.'''
self._size = 0

def __len__(self):
'''Return the number of elements in the stack.'''
return self._size

def IsEmpty(self):
'''Return True if the stack is empty'''
return  self._size == 0

def Push(self,e):
'''Add element e to the top of the Stack.'''
self._head = self._Node(e, self._head)      #create and link a new node
self._size +=1
self._ToList.append(e)

def Top(self):
'''Return (but do not remove) the element at the top of the stack.
Raise exception if the stack is empty
'''

if self.IsEmpty():
raise Exception('Stack is empty')
return  self._head._element             #top of stack is at head of list

def Pop(self):
'''Remove and return the element from the top of the stack (i.e. LIFO).
Raise exception if the stack is empty
'''
if self.IsEmpty():
raise Exception('Stack is empty')
self._head = self._head._next       #bypass the former top node
self._size -=1

def Count(self):
'''Return how many nodes the stack has'''
return self.__len__()

def Clear(self):
'''Delete all nodes'''
for i in range(self.Count()):
self.Pop()

def ToList(self):
return self._ToList

# Nested Node Class
class Node:
def __init__(self, element, next):
self.__element = element
self.__next = next

def get_next(self):
return self.__next

def get_element(self):
return self.__element

def __init__(self):
self.size = 0
self.data = []

def __len__(self):
return self.size

def __str__(self):
return str(self.data)

def is_empty(self):
return self.size == 0

def push(self, e):
self.size += 1

def top(self):
if self.is_empty():
raise Empty('Stack is empty')

def pop(self):
if self.is_empty():
raise Empty('Stack is empty')
self.size -= 1

Usage

x.push(10)
x.push(25)
x.push(55)

for i in range(x.size - 1, -1, -1):

print '|', x.data[i].get_element(), '|' ,
#next object

if x.data[i].get_next() == None:
print '--> None'
else:
print  x.data[i].get_next().get_element(), '-|---->  ',

Output

| 55 | 25 -|---->   | 25 | 10 -|---->   | 10 | --> None
class Node(object):
def __init__(self, data=None, next=None):
self.data = data
self.next = next

def setData(self, data):
self.data = data
return self.data

def setNext(self, next):
self.next = next

def getNext(self):
return self.next

def hasNext(self):
return self.next != None

def __init__(self):

def isEmpty(self):
return self.head == None

def insertAtBeginning(self, data):
newNode = Node()
newNode.setData(data)

if self.listLength() == 0:
else:

def insertAtEnd(self, data):
newNode = Node()
newNode.setData(data)

while current.getNext() != None:
current = current.getNext()

current.setNext(newNode)

def listLength(self):
count = 0

while current != None:
count += 1
current = current.getNext()
return count

def print_llist(self):
print("List Start.")
while current != None:
print(current.getData())
current = current.getNext()

print("List End.")

if __name__ == '__main__':
ll.insertAtBeginning(55)
ll.insertAtEnd(56)
ll.print_llist()
print(ll.listLength())

Here is my simple implementation:

class Node:
def __init__(self):
self.data = None
self.next = None
def __str__(self):
return "Data %s: Next -> %s"%(self.data, self.next)

def __init__(self):
def insertNode(self, data):
node = Node()
node.data = data
node.next = None
if self.head.data == None:
self.curNode = node
else:
self.curNode.next = node
self.curNode = node
def printList(self):

l.insertNode(1)
l.insertNode(2)
l.insertNode(34)

Output:

Data 1: Next -> Data 2: Next -> Data 34: Next -> Data 4: Next -> None

# llist — Linked list datatypes for Python

llist module implements linked list data structures. It supports a doubly linked list, i.e. dllist and a singly linked data structure sllist.

## dllist objects

This object represents a doubly linked list data structure.

### first

First dllistnode object in the list. None if list is empty.

### last

Last dllistnode object in the list. None if list is empty.

dllist objects also support the following methods:

### append(x)

Add x to the right side of the list and return inserted dllistnode.

### appendleft(x)

Add x to the left side of the list and return inserted dllistnode.

### appendright(x)

Add x to the right side of the list and return inserted dllistnode.

### clear()

Remove all nodes from the list.

### extend(iterable)

Append elements from iterable to the right side of the list.

### extendleft(iterable)

Append elements from iterable to the left side of the list.

### extendright(iterable)

Append elements from iterable to the right side of the list.

### insert(x[, before])

Add x to the right side of the list if before is not specified, or insert x to the left side of dllistnode before. Return inserted dllistnode.

### nodeat(index)

Return node (of type dllistnode) at index.

### pop()

Remove and return an element’s value from the right side of the list.

### popleft()

Remove and return an element’s value from the left side of the list.

### popright()

Remove and return an element’s value from the right side of the list

### remove(node)

Remove node from the list and return the element which was stored in it.

## dllistnode objects

### class llist.dllistnode([value])

Return a new doubly linked list node, initialized (optionally) with value.

### next

Next node in the list. This attribute is read-only.

### prev

Previous node in the list. This attribute is read-only.

### value

Value stored in this node.
Compiled from this reference

# sllist

class llist.sllist([iterable])
Return a new singly linked list initialized with elements from iterable. If iterable is not specified, the new sllist is empty.

A similar set of attributes and operations are defined for this sllist object. See this reference for more information.

Sample of a doubly linked list (save as linkedlist.py):

class node:
def __init__(self, before=None, cargo=None, next=None):
self._previous = before
self._cargo = cargo
self._next  = next

def __str__(self):
return str(self._cargo) or None

def __init__(self):
self._length = 0

n = node(None, cargo, self._head)
self._length += 1

def search(self,cargo):
while (node and node._cargo != cargo):
node = node._next
return node

def delete(self,cargo):
node = self.search(cargo)
if node:
prev = node._previous
nx = node._next
if prev:
prev._next = node._next
else:
nx._previous = None
if nx:
nx._previous = prev
else:
prev._next = None
self._length -= 1

def __str__(self):
print 'Size of linked list: ',self._length
while node:
print node
node = node._next

Testing (save as test.py):

def test():

print 'Testing Linked List'

print 'Linked List after insert nodes:'
l.__str__()

print 'Search some value, 30:'
node = l.search(30)
print node

print 'Delete some value, 30:'
node = l.delete(30)
l.__str__()

print 'Delete first element, 60:'
node = l.delete(60)
l.__str__()

print 'Delete last element, 10:'
node = l.delete(10)
l.__str__()

if __name__ == "__main__":
test()

Output:

Linked List after insert nodes:
Size of linked list:  6
60
50
40
30
20
10
Search some value, 30:
30
Delete some value, 30:
Size of linked list:  5
60
50
40
20
10
Delete first element, 60:
Size of linked list:  4
50
40
20
10
Delete last element, 10:
Size of linked list:  3
50
40
20

Here is my solution:

Implementation

class Node:
def __init__(self, initdata):
self.data = initdata
self.next = None

def get_data(self):
return self.data

def set_data(self, data):
self.data = data

def get_next(self):
return self.next

def set_next(self, node):
self.next = node

# ------------------------ Link List class ------------------------------- #

def __init__(self):

def is_empty(self):
return self.head == None

def traversal(self, data=None):
index = 0
found = False
while node is not None and not found:
if node.get_data() == data:
found = True
else:
node = node.get_next()
index += 1
return (node, index)

def size(self):
_, count = self.traversal(None)
return count

def search(self, data):
node, _ = self.traversal(data)
return node

node = Node(data)

def remove(self, data):
previous_node = None
found = False
while current_node is not None and not found:
if current_node.get_data() == data:
found = True
if previous_node:
previous_node.set_next(current_node.get_next())
else:
else:
previous_node = current_node
current_node = current_node.get_next()
return found

Usage

Original Implementation Idea

my double Linked List might be understandable to noobies.
If you are familiar with DS in C, this is quite readable.

class node:
def __init__(self):           ##Cluster of Nodes' properties
self.data=None
self.next=None
self.prev=None

def __init__(self):
self.t = node()                    // for future use
self.cur_node = node()             // current node
self.start=node()

def add(self,data):                          // appending the LL

self.new_node = node()
self.new_node.data=data
if self.cur_node.data is None:
self.start=self.new_node               //For the 1st node only

self.cur_node.next=self.new_node
self.new_node.prev=self.cur_node
self.cur_node=self.new_node

def backward_display(self):                  //Displays LL backwards
self.t=self.cur_node
while self.t.data is not None:
print(self.t.data)
self.t=self.t.prev

def forward_display(self):                   //Displays LL Forward
self.t=self.start
while self.t.data is not None:
print(self.t.data)
self.t=self.t.next
if self.t.next is None:
print(self.t.data)
break

def main(self):                          //This is kind of the main
function in C
ch=0
while ch is not 4:                    //Switch-case in C
if ch is 1:
data=int(input("Enter data to be added:"))
ll.main()
elif ch is 2:
ll.forward_display()
ll.main()
elif ch is 3:
ll.backward_display()
ll.main()
else:
print("Program ends!!")
return

ll.main()

Though many more simplifications can be added to this code, I thought a raw implementation would me more grabbable.

I did also write a Single Linked List based on some tutorial, which has the basic two Node and Linked List classes, and some additional methods for insertion, delete, reverse, sorting, and such.

It’s not the best or easiest, works OK though.

"""

Single Linked List (SLL):
A simple object-oriented implementation of Single Linked List (SLL)
with some associated methods, such as create list, count nodes, delete nodes, and such.

"""

class Node:
"""
Instantiates a node
"""
def __init__(self, value):
"""
Node class constructor which sets the value and link of the node

"""
self.info = value

"""
Instantiates the SLL class
"""
def __init__(self):
"""
SLL class constructor which sets the value and link of the node

"""
self.start = None

"""
Reads values from stdin and appends them to this list and creates a SLL with integer nodes

"""
try:
number_of_nodes = int(input("    Enter a positive integer between 1-50 for the number of nodes you wish to have in the list: "))
if number_of_nodes <= 0 or number_of_nodes > 51:
print("  The number of nodes though must be an integer between 1 to 50!")

except Exception as e:
print("  Error: ", e)

try:
for _ in range(number_of_nodes):
try:
data = int(input("    Enter an integer for the node to be inserted: "))
self.insert_node_at_end(data)
except Exception as e:
print("  Error: ", e)
except Exception as e:
print("  Error: ", e)

def count_sll_nodes(self):
"""
Counts the nodes of the linked list

"""
try:
p = self.start
n = 0
while p is not None:
n += 1

if n >= 1:
print(f"  The number of nodes in the linked list is {n}")
else:
print(f"  The SLL does not have a node!")
except Exception as e:
print("  Error: ", e)

def search_sll_nodes(self, x):
"""
Searches the x integer in the linked list
"""
try:
position =  1
p = self.start
while p is not None:
if p.info == x:
print(f"  YAAAY! We found {x} at position {position}")
return True

#Increment the position
position += 1
#Assign the next node to the current node
else:
print(f"  Sorry! We couldn't find {x} at any position. Maybe, you might want to use option 9 and try again later!")
return False
except Exception as e:
print("  Error: ", e)

def display_sll(self):
"""
Displays the list
"""
try:
if self.start is None:
print("  Single linked list is empty!")
return

display_sll = "  Single linked list nodes are: "
p = self.start
while p is not None:
display_sll += str(p.info) + "t"

print(display_sll)

except Exception as e:
print("  Error: ", e)

def insert_node_in_beginning(self, data):
"""
Inserts an integer in the beginning of the linked list

"""
try:
temp = Node(data)
self.start = temp
except Exception as e:
print("  Error: ", e)

def insert_node_at_end(self, data):
"""
Inserts an integer at the end of the linked list

"""
try:
temp = Node(data)
if self.start is None:
self.start = temp
return

p = self.start
while p.link is not None:
except Exception as e:
print("  Error: ", e)

def insert_node_after_another(self, data, x):
"""
Inserts an integer after the x node

"""
try:
p = self.start

while p is not None:
if p.info == x:
break

if p is None:
print(f"  Sorry! {x} is not in the list.")
else:
temp = Node(data)
except Exception as e:
print("  Error: ", e)

def insert_node_before_another(self, data, x):
"""
Inserts an integer before the x node

"""

try:

# If list is empty
if self.start is None:
print("  Sorry! The list is empty.")
return
# If x is the first node, and new node should be inserted before the first node
if x == self.start.info:
temp = Node(data)

# Finding the reference to the prior node containing x
p = self.start
while p.link is not None:
if p.link.info == x:
break

if p.link is not None:
print(f"  Sorry! {x} is not in the list.")
else:
temp = Node(data)

except Exception as e:
print("  Error: ", e)

def insert_node_at_position(self, data, k):
"""
Inserts an integer in k position of the linked list

"""
try:
# if we wish to insert at the first node
if k == 1:
temp = Node(data)
self.start = temp
return

p = self.start
i = 1

while i < k-1 and p is not None:
i += 1

if p is None:
print(f"  The max position is {i}")
else:
temp = Node(data)
self.start = temp

except Exception as e:
print("  Error: ", e)

def delete_a_node(self, x):
"""
Deletes a node of a linked list

"""
try:
# If list is empty
if self.start is None:
print("  Sorry! The list is empty.")
return

# If there is only one node
if self.start.info == x:

# If more than one node exists
p = self.start
while p.link is not None:
if p.link.info == x:
break

if p.link is None:
print(f"  Sorry! {x} is not in the list.")
else:

except Exception as e:
print("  Error: ", e)

def delete_sll_first_node(self):
"""
Deletes the first node of a linked list

"""
try:
if self.start is None:
return

except Exception as e:
print("  Error: ", e)

def delete_sll_last_node(self):
"""
Deletes the last node of a linked list

"""
try:

# If the list is empty
if self.start is None:
return

# If there is only one node
if self.start.link is None:
self.start = None
return

# If there is more than one node
p = self.start

# Increment until we find the node prior to the last node

except Exception as e:
print("  Error: ", e)

def reverse_sll(self):
"""
Reverses the linked list

"""

try:

prev = None
p = self.start
while p is not None:
prev = p
p = next
self.start = prev

except Exception as e:
print("  Error: ", e)

def bubble_sort_sll_nodes_data(self):
"""
Bubble sorts the linked list on integer values

"""

try:

# If the list is empty or there is only one node
if self.start is None or self.start.link is None:
print("  The list has no or only one node and sorting is not required.")
end = None

while end != self.start.link:
p = self.start
while p.link != end:
if p.info > q.info:
p.info, q.info = q.info, p.info
end = p

except Exception as e:
print("  Error: ", e)

def bubble_sort_sll(self):
"""
Bubble sorts the linked list

"""

try:

# If the list is empty or there is only one node
if self.start is None or self.start.link is None:
print("  The list has no or only one node and sorting is not required.")
end = None

while end != self.start.link:
r = p = self.start
while p.link != end:
if p.info > q.info:
if  p != self.start:
else:
self.start = q
p, q = q, p
r = p
end = p

except Exception as e:
print("  Error: ", e)

def sll_has_cycle(self):
"""
Tests the list for cycles using Tortoise and Hare Algorithm (Floyd's cycle detection algorithm)
"""

try:

if self.find_sll_cycle() is None:
return False
else:
return True

except Exception as e:
print("  Error: ", e)

def find_sll_cycle(self):
"""
Finds cycles in the list, if any
"""

try:

# If there is one node or none, there is no cycle
if self.start is None or self.start.link is None:
return None

# Otherwise,
slowR = self.start
fastR = self.start

while slowR is not None and fastR is not None:
if slowR == fastR:
return slowR

return None

except Exception as e:
print("  Error: ", e)

def remove_cycle_from_sll(self):
"""
Removes the cycles
"""

try:

c = self.find_sll_cycle()

# If there is no cycle
if c is None:
return

print(f"  There is a cycle at node: ", c.info)

p = c
q = c
len_cycles = 0
while True:
len_cycles += 1

if p == q:
break

print(f"  The cycle length is {len_cycles}")

len_rem_list = 0
p = self.start

while p != q:
len_rem_list += 1

print(f"  The number of nodes not included in the cycle is {len_rem_list}")

length_list = len_rem_list + len_cycles

print(f"  The SLL length is {length_list}")

# This for loop goes to the end of the SLL, and set the last node to None and the cycle is removed.
p = self.start
for _ in range(length_list-1):

except Exception as e:
print("  Error: ", e)

def insert_cycle_in_sll(self, x):
"""
Inserts a cycle at a node that contains x

"""

try:

if self.start is None:
return False

p = self.start
px = None
prev = None

while p is not None:
if p.info == x:
px = p
prev = p

if px is not None:
else:
print(f"  Sorry! {x} is not in the list.")

except Exception as e:
print("  Error: ", e)

def merge_using_new_list(self, list2):
"""
Merges two already sorted SLLs by creating new lists
"""
merge_list.start = self._merge_using_new_list(self.start, list2.start)
return merge_list

def _merge_using_new_list(self, p1, p2):
"""
Private method of merge_using_new_list
"""
if p1.info <= p2.info:
Start_merge = Node(p1.info)
else:
Start_merge = Node(p2.info)
pM = Start_merge

while p1 is not None and p2 is not None:
if p1.info <= p2.info:
else:

#If the second list is finished, yet the first list has some nodes
while p1 is not None:

#If the second list is finished, yet the first list has some nodes
while p2 is not None:

return Start_merge

def merge_inplace(self, list2):
"""
Merges two already sorted SLLs in place in O(1) of space
"""
merge_list.start = self._merge_inplace(self.start, list2.start)
return merge_list

def _merge_inplace(self, p1, p2):
"""
Merges two already sorted SLLs in place in O(1) of space
"""
if p1.info <= p2.info:
Start_merge = p1
else:
Start_merge = p2
pM = Start_merge

while p1 is not None and p2 is not None:
if p1.info <= p2.info:
else:

if p1 is None:
else:

return Start_merge

def merge_sort_sll(self):
"""
Sorts the linked list using merge sort algorithm
"""
self.start = self._merge_sort_recursive(self.start)

def _merge_sort_recursive(self, list_start):
"""
Recursively calls the merge sort algorithm for two divided lists
"""

# If the list is empty or has only one node
if list_start is None or list_start.link is None:
return list_start

# If the list has two nodes or more
start_one = list_start
start_two = self._divide_list(self_start)
start_one = self._merge_sort_recursive(start_one)
start_two = self._merge_sort_recursive(start_two)
start_merge = self._merge_inplace(start_one, start_two)

return start_merge

def _divide_list(self, p):
"""
Divides the linked list into two almost equally sized lists
"""

# Refers to the third nodes of the list

while q is not None and p is not None:
# Increments p one node at the time
# Increments q two nodes at the time

return start_two

def concat_second_list_to_sll(self, list2):
"""
Concatenates another SLL to an existing SLL
"""

# If the second SLL has no node
if list2.start is None:
return

# If the original SLL has no node
if self.start is None:
self.start = list2.start
return

# Otherwise traverse the original SLL
p = self.start
while p.link is not None:

# Link the last node to the first node of the second SLL

def test_merge_using_new_list_and_inplace(self):
"""

"""

print("1️⃣  The unsorted first list is: ")
LIST_ONE.display_sll()

print("2️⃣  The unsorted second list is: ")
LIST_TWO.display_sll()

LIST_ONE.bubble_sort_sll_nodes_data()
LIST_TWO.bubble_sort_sll_nodes_data()

print("1️⃣  The sorted first list is: ")
LIST_ONE.display_sll()

print("2️⃣  The sorted second list is: ")
LIST_TWO.display_sll()

LIST_THREE = LIST_ONE.merge_using_new_list(LIST_TWO)

print("The merged list by creating a new list is: ")
LIST_THREE.display_sll()

LIST_FOUR = LIST_ONE.merge_inplace(LIST_TWO)

print("The in-place merged list is: ")
LIST_FOUR.display_sll()

def test_all_methods(self):
"""
Tests all methods of the SLL class
"""

OPTIONS_HELP = """

Select a method from 1-19:

ℹ️   (1)       Create a single liked list (SLL).
ℹ️   (2)       Display the SLL.
ℹ️   (3)       Count the nodes of SLL.
ℹ️   (4)       Search the SLL.
ℹ️   (5)       Insert a node at the beginning of the SLL.
ℹ️   (6)       Insert a node at the end of the SLL.
ℹ️   (7)       Insert a node after a specified node of the SLL.
ℹ️   (8)       Insert a node before a specified node of the SLL.
ℹ️   (9)       Delete the first node of SLL.
ℹ️   (10)      Delete the last node of the SLL.
ℹ️   (11)      Delete a node you wish to remove.
ℹ️   (12)      Reverse the SLL.
ℹ️   (13)      Bubble sort the SLL by only exchanging the integer values.
ℹ️   (14)      Bubble sort the SLL by exchanging links.
ℹ️   (15)      Merge sort the SLL.
ℹ️   (16)      Insert a cycle in the SLL.
ℹ️   (17)      Detect if the SLL has a cycle.
ℹ️   (18)      Remove cycle in the SLL.
ℹ️   (19)      Test merging two bubble-sorted SLLs.
ℹ️   (20)      Concatenate a second list to the SLL.
ℹ️   (21)      Exit.

"""

while True:

print(OPTIONS_HELP)

UI_OPTION = int(input("    Enter an integer for the method you wish to run using the above help: "))

if UI_OPTION == 1:
data = int(input("    Enter an integer to be inserted at the end of the list: "))
x = int(input("    Enter an integer to be inserted after that: "))
self.insert_node_after_another(data, x)
elif UI_OPTION == 2:
self.display_sll()
elif UI_OPTION == 3:
self.count_sll_nodes()
elif UI_OPTION == 4:
data = int(input("    Enter an integer to be searched: "))
self.search_sll_nodes(data)
elif UI_OPTION == 5:
data = int(input("    Enter an integer to be inserted at the beginning: "))
self.insert_node_in_beginning(data)
elif UI_OPTION == 6:
data = int(input("    Enter an integer to be inserted at the end: "))
self.insert_node_at_end(data)
elif UI_OPTION == 7:
data = int(input("    Enter an integer to be inserted: "))
x = int(input("    Enter an integer to be inserted before that: "))
self.insert_node_before_another(data, x)
elif UI_OPTION == 8:
data = int(input("    Enter an integer for the node to be inserted: "))
k = int(input("    Enter an integer for the position at which you wish to insert the node: "))
self.insert_node_before_another(data, k)
elif UI_OPTION == 9:
self.delete_sll_first_node()
elif UI_OPTION == 10:
self.delete_sll_last_node()
elif UI_OPTION == 11:
data = int(input("    Enter an integer for the node you wish to remove: "))
self.delete_a_node(data)
elif UI_OPTION == 12:
self.reverse_sll()
elif UI_OPTION == 13:
self.bubble_sort_sll_nodes_data()
elif UI_OPTION == 14:
self.bubble_sort_sll()
elif UI_OPTION == 15:
self.merge_sort_sll()
elif UI_OPTION == 16:
data = int(input("    Enter an integer at which a cycle has to be formed: "))
self.insert_cycle_in_sll(data)
elif UI_OPTION == 17:
if self.sll_has_cycle():
print("  The linked list has a cycle. ")
else:
print("  YAAAY! The linked list does not have a cycle. ")
elif UI_OPTION == 18:
self.remove_cycle_from_sll()
elif UI_OPTION == 19:
self.test_merge_using_new_list_and_inplace()
elif UI_OPTION == 20:
self.concat_second_list_to_sll(list2)
elif UI_OPTION == 21:
break
else:
print("  Option must be an integer, between 1 to 21.")

print()

if __name__ == '__main__':
# Instantiates a new SLL object
SLL_OBJECT.test_all_methods()

Expanding Nick Stinemates’s answer

class Node(object):
def __init__(self):
self.data = None
self.next = None

def __init__(self):

def prepend_node(self, data):
new_node = Node()
new_node.data = data

def append_node(self, data):
new_node = Node()
new_node.data = data
while current.next:
current = current.next
current.next = new_node

def reverse(self):
""" In-place reversal, modifies exiting list"""
previous = None

while current_node:
temp =  current_node.next
current_node.next = previous
previous = current_node
current_node = temp

def search(self, data):
try:
while current_node.data != data:
current_node = current_node.next
return True
except:
return False

def display(self):
if self.head is None:
print("Linked list is empty")
else:
while current_node:
print(current_node.data)
current_node = current_node.next

def list_length(self):
list_length = 0
while current_node:
list_length += 1
current_node = current_node.next
return list_length

def main():

print("Lenght of singly linked list is: " + str(linked_list.list_length()))

if __name__ == "__main__":
main()

Current Implementation of Linked List in Python requires for creation of a separate class, called Node, so that they can be connected using a main Linked List class. In the provided implementation, the Linked List is created without defining a separate class for a node. Using the proposed implementation, Linked Lists are easier to understand and can be simply visualized using the print function.

def __init__(self):
self.outer = None

self.outer = [dt, self.outer]

p = self.outer
if not p:
self.outer = [dt, None]
return
while p[1]:
p = p[1]
p[1] = [dt, None]

def visualize(self):
p = self.outer
l = 'Linkedlist: '
while p:
l += (str(p[0])+'->')
p = p[1]
print(l + 'None')