The LinkedList class provides a singly linked list implementation where each node points to the next node.
Node A singly linked list node implementation.
Constructor from dsa.singlylinkedlist import Node node = Node( 10 ) print (node.value) # Output: 10 print (node.next) # Output: None
Properties The value stored in the node
Reference to the next node in the list
LinkedList A singly linked list implementation.
Constructor head
Node | None
default: "None"
Reference to the head node
tail
Node | None
default: "None"
Reference to the tail node
The number of nodes in the linked list
If only the head node is specified, tail is set to the head node and count is automatically set to 1. If both head and tail nodes are specified, count should be specified as well.
from dsa.singlylinkedlist import LinkedList, Node # Create an empty linked list ll = LinkedList() # Create with a single node node = Node( 10 ) ll = LinkedList( head = node) # Create with head and tail head = Node( 1 ) tail = Node( 3 ) head.next = Node( 2 ) head.next.next = tail ll = LinkedList( head = head, tail = tail, count = 3 )
Properties Reference to the first node in the list
Reference to the last node in the list
The number of nodes in the linked list
Methods from_list(mylist) Class method to create a linked list from a list.
A list or container to convert from
Returns: A linked list containing the items from mylistll = LinkedList.from_list([ 1 , 2 , 3 , 4 , 5 ]) print (ll.to_list()) # Output: [1, 2, 3, 4, 5]
to_list() Convert the linked list to a Python list.
Returns: List with contents of linked listll = LinkedList.from_list([ 1 , 2 , 3 ]) print (ll.to_list()) # Output: [1, 2, 3]
traverse() Print the contents of the linked list.
ll = LinkedList.from_list([ 1 , 2 , 3 , 4 , 5 ]) ll.traverse() # Output: 1 2 3 4 5
search(value) Search for a value in the linked list.
Returns: The Node containing the value, or None if not foundll = LinkedList.from_list([ 10 , 20 , 30 ]) node = ll.search( 20 ) if node: print ( f "Found: { node.value } " ) # Output: Found: 20 else : print ( "Not found" ) node = ll.search( 99 ) print (node) # Output: None
is_empty() Check if the linked list is empty.
Returns: True if the list is empty, False otherwisell = LinkedList() print (ll.is_empty()) # Output: True ll.append( 1 ) print (ll.is_empty()) # Output: False
prepend(value) Place a value at the beginning of the linked list.
Returns: Nonell = LinkedList.from_list([ 2 , 3 , 4 ]) ll.prepend( 1 ) print (ll.to_list()) # Output: [1, 2, 3, 4]
append(value) Place a value at the end of the linked list.
Returns: Nonell = LinkedList.from_list([ 1 , 2 , 3 ]) ll.append( 4 ) print (ll.to_list()) # Output: [1, 2, 3, 4]
insert_after(after_value, value) Insert a value after a specified value.
The value to insert after
Returns: NoneRaises:
ValueError: If the after_value is not found
ll = LinkedList.from_list([ 1 , 2 , 4 ]) ll.insert_after( 2 , 3 ) print (ll.to_list()) # Output: [1, 2, 3, 4] # ll.insert_after(10, 5) # Raises ValueError: "Value not found"
delete(value) Delete the first occurrence of a value in the linked list.
Returns: NoneRaises:
ValueError: If the value is not found
ll = LinkedList.from_list([ 1 , 2 , 3 , 2 , 4 ]) ll.delete( 2 ) # Deletes first occurrence print (ll.to_list()) # Output: [1, 3, 2, 4] # ll.delete(10) # Raises ValueError: "Value not found"
delete_head() Delete the head node in the linked list.
Returns: NoneRaises:
IndexError: If linked list is empty
ll = LinkedList.from_list([ 1 , 2 , 3 ]) ll.delete_head() print (ll.to_list()) # Output: [2, 3] # empty_ll = LinkedList() # empty_ll.delete_head() # Raises IndexError: "LinkedList is Empty"
delete_tail() Delete the last node in the linked list.
Returns: NoneRaises:
IndexError: If linked list is empty
ll = LinkedList.from_list([ 1 , 2 , 3 ]) ll.delete_tail() print (ll.to_list()) # Output: [1, 2]
Special Methods __getitem__(index) Return value at a specified index.
Returns: The value at the specified indexRaises:
IndexError: If index is out of bounds
ll = LinkedList.from_list([ 10 , 20 , 30 , 40 ]) print (ll[ 0 ]) # Output: 10 print (ll[ 2 ]) # Output: 30 # print(ll[10]) # Raises IndexError: "Index Out of Bounds"
__len__() Return the number of elements in the linked list.
ll = LinkedList.from_list([ 1 , 2 , 3 , 4 , 5 ]) print ( len (ll)) # Output: 5
__eq__(other) Compare two LinkedList objects for value-based equality.
Returns: True if both are LinkedList instances with equal contentsll1 = LinkedList.from_list([ 1 , 2 , 3 ]) ll2 = LinkedList.from_list([ 1 , 2 , 3 ]) ll3 = LinkedList.from_list([ 1 , 2 , 4 ]) print (ll1 == ll2) # Output: True print (ll1 == ll3) # Output: False
__repr__() Return a string representation of the linked list.
ll = LinkedList.from_list([ 1 , 2 , 3 ]) print (ll) # Output: [ 1 2 3 ] Count: 3
Complete Example
Working with SinglyLinkedList
from dsa.singlylinkedlist import LinkedList # Create a linked list from a list ll = LinkedList.from_list([ 10 , 20 , 30 , 40 , 50 ]) # Display the list print ( f "Initial list: { ll.to_list() } " ) print ( f "Length: { len (ll) } " ) # Access elements by index print ( f "First element: { ll[ 0 ] } " ) print ( f "Third element: { ll[ 2 ] } " ) # Add elements ll.prepend( 5 ) ll.append( 60 ) ll.insert_after( 30 , 35 ) print ( f "After additions: { ll.to_list() } " ) # Output: [5, 10, 20, 30, 35, 40, 50, 60] # Search for a value node = ll.search( 35 ) if node: print ( f "Found { node.value } at { node } " ) # Delete elements ll.delete( 35 ) ll.delete_head() ll.delete_tail() print ( f "After deletions: { ll.to_list() } " ) # Output: [10, 20, 30, 40, 50] # Traverse and print print ( "Traversal: " , end = "" ) ll.traverse() # Output: 10 20 30 40 50 # Check if empty print ( f "Is empty: { ll.is_empty() } " ) # Compare lists ll2 = LinkedList.from_list([ 10 , 20 , 30 , 40 , 50 ]) print ( f "Lists equal: { ll == ll2 } " ) # Output: True # Clear the list while not ll.is_empty(): ll.delete_head() print ( f "After clearing, is empty: { ll.is_empty() } " ) # Output: True
Use Cases Example: Reverse a linked list
Example: Find middle element
Example: Remove duplicates
def reverse_list ( ll ): values = ll.to_list() values.reverse() return LinkedList.from_list(values) ll = LinkedList.from_list([ 1 , 2 , 3 , 4 , 5 ]) reversed_ll = reverse_list(ll) print (reversed_ll.to_list()) # Output: [5, 4, 3, 2, 1]