Linked List in Python
The complete code can be found at linked_list.py.
A few notes on implementation
When translating the pseudo code into its Python's implementation, we shall pay attention to comparison between objects.
def remove_first(self):
if self.is_empty():
raise IndexError('Remove from empty linked list')
else:
self._head = self._head.next
if self._head is None:
self._tail = None
self._size -= 1
The code above shows the procedure of removing the first element in a linked list. If there was only one element, then both head
and tail
should be set to None
. To check whether current head is None
, we use the built-in is
, instead of ==
.
is
will return True if two variables point to the same object (in memory).==
will return True if the objects referred to by the variables are equal. See more at Is there a difference between "==" and "is"?.
Shorter yet obscure
As for nodes in linked lists, we shall focus on how to assign values and relink the next
pointer. Sometimes, relinking twice can be shortened by leveraging of __init__
. For example, the following code is to add an element onto a stack (we will check it later):
# push() = add_first()
def push(self, item):
node = Node(item)
node.next = self._head
self._head = node
self._size += 1
It can be shortened to:
# push() = add_first()
def push(self, item):
self._head = Node(item, self._head)
self._size += 1
It is shorter, but a bit more difficult to understand. Some people may strive for elegance and prefer to shorter code, while others may give a priority to ease-of-understanding. Again, it is a trade-off.
Stack based on linked lists
In this subsection, we are going to implement a stack based on singly linked lists.
A stack is a collection of objects that are inserted and removed according to the last-in, first-out (LIFO) policy.
The two main operations of a stack can be described in the following, and both of them run in \(O(1)\):
push()
:add_first()
in a linked listpop()
:remove_first()
in a linked list
As we can see, only the head
pointer is necessary here. The complete code can be found at linked_stack.py, which is a simplified implementation of a linked list.
Array-based stack | LinkedList-based stack | Note | |
---|---|---|---|
push() | \(O(1)\) | \(O(1)\) | Array-based is amortized |
pop() | \(O(1)\) | \(O(1)\) | Array-based is amortized |
Queue based on linked lists
In this subsection, we are going to implement a stack based on singly linked lists.
A queue is a collection of objects that are inserted and removed according to the first-in, first-out (FIFO) policy.
The two main operations of a queue can be described in the following, and both of them run in \(O(1)\):
enqueue()
:add_last()
in a linked listdequeue()
:remove_first()
in a linked list
We shall maintain both head
and tail
for a queue. The complete code can be found at linked_queue.py, which is also a simplified implementation of a linked list.
Array-based queue | LinkedList-based queue | Note | |
---|---|---|---|
enqueue() | \(O(1)\) | \(O(1)\) | Array-based is amortized |
dequeue() | \(O(N)\) | \(O(1)\) | Not the circular array |
Circularly linked list
A circularly linked list is essentially a singly linked list in which the next
reference to the tail node is set to refer back to the head of the list (rather than null
).
![](image/circular_linkedlist.png)
This ADT can be used in applications in which data can be more naturally viewed as having a cyclic order, with well-defined neighboring relationships, but no fixed beginning or end. Note that there is a link from tail
to head
, so only tail
is required for a circularly linked list.
In what follows, We elaborate some key implementations of this ADT in Python, and the full code can be found at circular_linked_list.py:
addFirst()
def add_first(self, item):
if self.is_empty():
self._tail = self.Node(item)
self._tail.next = self._tail
else:
self._tail.next = self.Node(item, self._tail.next)
self._size += 1
addLast()
def add_last(self, item):
self.add_first(item)
self._tail = self._tail.next
removeFirst()
def remove_first(self):
if self.is_empty():
raise NoElement
head = self._tail.next
if head is self._tail:
# only one element
self._tail = None
else:
self._tail.next = head.next
self._size -= 1