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To implement a singly linked list in Python, define a Node that stores a value and a reference to the next node, then define a list class that tracks at least head. Keep a tail reference when you need constant-time appends, and maintain a size counter if callers need the length without traversing the chain.
The implementation below is complete enough for learning and small specialized structures: it supports append, prepend, search, insertion after a known node, removal, iteration, length checks, and invariant validation. It also makes empty-list behavior explicit instead of leaving edge cases to chance.
How a linked list works
A singly linked list is a chain of nodes. Each node has two fields:
- value: the payload, such as an integer, string, or object.
- next: a reference to the next node, or
Nonefor the final node.
The list object stores a reference to the first node in head. A practical implementation also stores tail, pointing to the final node, and size, recording how many nodes are present.
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Unlike a Python list, a linked list does not keep elements in one contiguous array. To reach the fifth element, Python must follow four next references. That is why traversal, search, and index lookup are linear operations.
A complete singly linked-list implementation
This version uses the following policies: find returns a node or None; remove returns a Boolean; removing from an empty list returns False; and inserting after a node that is not in the list raises ValueError. Documenting these choices is important because empty operations otherwise tend to produce inconsistent APIs.
class Node:
def __init__(self, value, next_node=None):
self.value = value
self.next = next_node
def __repr__(self):
return f"Node({self.value!r})"
class LinkedList:
def __init__(self):
self.head = None
self.tail = None
self.size = 0
def __len__(self):
return self.size
def __iter__(self):
current = self.head
while current is not None:
yield current.value
current = current.next
def __contains__(self, value):
return self.find(value) is not None
def __repr__(self):
values = ", ".join(repr(value) for value in self)
return f"LinkedList([{values}])"
def append(self, value):
"""Add value at the end in O(1) time."""
node = Node(value)
if self.head is None:
self.head = self.tail = node
else:
self.tail.next = node
self.tail = node
self.size += 1
def prepend(self, value):
"""Add value at the beginning in O(1) time."""
node = Node(value, self.head)
self.head = node
if self.tail is None:
self.tail = node
self.size += 1
def find(self, value):
"""Return the first matching Node, or None."""
current = self.head
while current is not None:
if current.value == value:
return current
current = current.next
return None
def insert_after(self, node, value):
"""Insert value after node and return the new Node."""
if node is None:
raise ValueError("node must not be None")
current = self.head
while current is not node:
if current is None:
raise ValueError("node does not belong to this list")
current = current.next
new_node = Node(value, node.next)
node.next = new_node
if self.tail is node:
self.tail = new_node
self.size += 1
return new_node
def pop_front(self):
"""Remove and return the first value, or None when empty."""
if self.head is None:
return None
removed = self.head
self.head = removed.next
removed.next = None
self.size -= 1
if self.head is None:
self.tail = None
return removed.value
def remove(self, value):
"""Remove the first matching value and return True if removed."""
previous = None
current = self.head
while current is not None:
if current.value == value:
if previous is None:
self.head = current.next
else:
previous.next = current.next
if self.tail is current:
self.tail = previous
current.next = None
self.size -= 1
if self.size == 0:
self.head = self.tail = None
return True
previous, current = current, current.next
return False
def clear(self):
"""Detach every node and reset the list."""
current = self.head
while current is not None:
following = current.next
current.next = None
current = following
self.head = self.tail = None
self.size = 0
def check_invariants(self):
"""Raise AssertionError if internal links and metadata disagree."""
if self.size == 0:
assert self.head is None and self.tail is None
return True
assert self.head is not None and self.tail is not None
count = 0
current = self.head
last = None
while current is not None:
count += 1
last = current
current = current.next
assert count == self.size
assert last is self.tail
assert self.tail.next is None
return True
Save the code in a file such as linked_list.py. It uses only the Python standard language features and runs on current Python 3 versions.
Why the endpoint updates matter
Appending to an empty list must set both head and tail. Prepending to a non-empty list changes only head; prepending to an empty list must also initialize tail. Removing the only node must clear both references. Removing the tail must move tail to the predecessor and set its next to None. The size counter must change exactly once for every successful insertion or removal.
Why insert_after accepts a node
Once a caller already has a predecessor node, insertion requires changing two references: the new node points to the old successor, and the predecessor points to the new node. The pointer changes are constant time. This method still verifies membership by traversal, because accepting an arbitrary node from another list would corrupt this list’s chain.
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Using the class
from linked_list import LinkedList
numbers = LinkedList()
print(numbers.pop_front()) # None
numbers.append(10)
numbers.append(30)
numbers.prepend(5)
middle = numbers.find(10)
numbers.insert_after(middle, 20)
print(list(numbers)) # [5, 10, 20, 30]
print(len(numbers)) # 4
print(numbers.tail.value) # 30
print(numbers.remove(5)) # True
print(numbers.remove(999)) # False
print(list(numbers)) # [10, 20, 30]
numbers.check_invariants() # True
Iteration is implemented with a generator, so ordinary constructs such as for value in numbers, list(numbers), and membership checks work without exposing pointer manipulation to every caller.
Operation complexity
With both head and tail, the common operations have these asymptotic costs:
| Operation | Singly linked list | Python list |
collections.deque |
|---|---|---|---|
| Indexing | O(n) traversal | O(1) | O(1) at ends; slower in the middle |
| Prepend | O(1) | O(n) because elements shift | Approximately O(1) with appendleft |
| Append | O(1) with tail; O(n) without it |
Amortized O(1) | Approximately O(1) |
| Search | O(n) | O(n) | O(n) |
| Remove after predecessor is known | O(1) | Usually requires shifting | Endpoint operations are approximately O(1) |
The costs describe the number of link or element operations as the structure grows. They do not mean every operation takes the same wall-clock time: a Python node is a separate object, and pointer-heavy traversal can have more allocation and cache overhead than traversing a compact array.
Why search and indexing are linear
A linked list has no direct address calculation for position i. Starting at head, code must follow one link at a time until it reaches that position or the end. Keeping a size field makes len(numbers) constant time, but it does not make indexing or searching constant time.
Why a tail reference changes append
Without tail, append must walk from head to the final node, which is O(n). With tail, append sets tail.next, moves tail, and increments size, all in O(1). The trade-off is an additional invariant to maintain during every mutation.
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Linked list, Python list, or deque?
Choose based on the operations your workload actually performs rather than on the data structure’s name.
| Choose | Best fit | Reason | Main limitation |
|---|---|---|---|
| Custom linked list | Teaching links, node-based algorithms, or code that already holds node references | Insertion or removal after a known node changes a few references | Linear indexing, object overhead, and more invariants to maintain |
Python list |
Random access, compact storage, sorting, slicing, and cache-friendly iteration | Contiguous references provide O(1) indexing | Inserting or deleting near the front or middle shifts references |
collections.deque |
Production queues, stacks, and double-ended workloads | Designed for approximately O(1) appends and pops at either end | It is not intended to provide fast middle indexing or arbitrary node references |
Python’s tutorial recommends collections.deque for queues because it is designed for fast appends and pops from both ends. The collections documentation describes approximately O(1) performance in either direction for endpoint operations. CPython’s FAQ describes lists as variable-length arrays backed by a contiguous array of references, not linked lists.
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queue.append("second")
print(queue.popleft()) # first
Use the custom class when the educational or algorithmic purpose is the linked structure itself. For ordinary application queues, a deque normally avoids unnecessary Python-level node objects.
Testing the edge cases
Linked lists often fail at transitions between zero, one, and two nodes. Include tests for each mutation boundary:
def test_linked_list():
linked = LinkedList()
assert len(linked) == 0
assert linked.head is None and linked.tail is None
assert linked.pop_front() is None
assert linked.remove("missing") is False
linked.append("only")
assert linked.head is linked.tail
assert linked.pop_front() == "only"
assert len(linked) == 0
linked.check_invariants()
linked.append("a")
linked.append("b")
linked.prepend("start")
assert list(linked) == ["start", "a", "b"]
assert linked.remove("start") is True # head removal
assert linked.remove("b") is True # tail removal
assert list(linked) == ["a"]
assert linked.remove("a") is True # final node
linked.check_invariants()
linked.append(1)
linked.append(1)
assert linked.remove(1) is True # removes first duplicate only
assert list(linked) == [1]
linked.clear()
assert len(linked) == 0
linked.check_invariants()
test_linked_list()
- Empty list: verify
head,tail, andsizeagree. - One node: removing it must clear both endpoint references.
- Head removal: the second node becomes the head.
- Tail removal: the predecessor becomes the tail and its
nextisNone. - Duplicate values: decide whether removal means the first match, all matches, or a specific node; this implementation removes the first match.
- Repeated operations: run alternating appends, prepends, removals, and clears, checking invariants after each mutation.
Common implementation problems and fixes
Append becomes slow
Symptom: repeatedly appending takes progressively longer. Cause: the implementation searches for the final node every time. Fix: maintain tail and update it when appending, removing the tail, or removing the last remaining node.
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The list loses its last node
Symptom: iteration stops early after insertion. Cause: a new node was assigned to current.next before its old successor was saved. Fix: construct the new node with next_node=current.next, then assign current.next = new_node.
Tail points into the middle
Symptom: later appends overwrite or bypass nodes. Cause: removing the tail did not move tail to its predecessor. Fix: track previous during traversal and assign self.tail = previous when the removed node is the tail.
An infinite loop appears during iteration
Symptom: a loop never reaches None. Cause: a link points back to an earlier node, often because of accidental reuse of a node or a faulty insertion. Fix: run check_invariants, inspect each next assignment, and add cycle detection if your application permits externally supplied nodes.
Removing by value is unexpectedly expensive
Symptom: remove(value) is slow for large lists. Cause: finding the value is O(n), even though changing links after the predecessor is known is O(1). Fix: retain the node and its predecessor when your algorithm already has them, or use a different structure when fast lookup is the primary requirement.
Useful extensions
Index-based access
You can add get(index) by rejecting negative or out-of-range indexes and traversing until the requested position. Its time complexity remains O(n). Do not present it as equivalent to Python-list indexing.
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Doubly linked lists
A doubly linked node adds a prev reference. That enables backward traversal and simpler removal when you already hold a node, but every insertion and deletion must update both directions. It also consumes more memory and creates more invariants.
Circular linked lists
A circular list points the final node back to the head instead of using None as a terminator. Traversal must stop after a known number of nodes or when it returns to the starting node; ordinary while current is not None loops are unsafe.
Memory and ownership
Each node is a Python object, so a custom linked list generally uses more memory than a Python list holding the same values. If nodes are exposed to callers, document who may mutate next; unrestricted external mutation can invalidate tail and size. Keeping mutation methods on the list object and validating invariants during tests is safer.
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Final decision guide
- Implement a custom linked list when learning pointer updates, experimenting with node-based algorithms, or operating on node references you already hold.
- Use a Python
listwhen indexing, slicing, compact storage, and iteration speed matter. - Use
collections.dequefor production queues, stacks, and operations at both ends. - Regardless of the structure, test empty, single-node, endpoint-removal, duplicate-value, and repeated-mutation cases.
Frequently Asked Questions
Can I make a linked list support negative indexes like a Python list?
Yes, but you must define the behavior yourself. A negative index requires knowing the length and then traversing to the corresponding node, so it remains O(n); it does not gain Python-list-style constant-time access.
How can I reverse a singly linked list in place?
Walk through the chain while retaining the next node, point the current node backward to the previous node, then advance both pointers. At the end, make the old tail the new head and the old head the new tail.
Should node attributes be public?
Public attributes keep a teaching implementation clear. In production code, restrict mutation or expose methods if callers changing next directly could break endpoint and size invariants.
What is the safest way to detect a cycle?
Use two pointers moving at different speeds (a slow pointer advancing one node and a fast pointer advancing two). If they meet, a cycle exists; if the fast pointer reaches None, the chain is acyclic.
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