Python Program For Removing Duplicates From A Sorted Linked List

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Write a function that takes a list sorted in non-decreasing order and deletes any duplicate nodes from the list. The list should only be traversed once.
For example if the linked list is 11->11->11->21->43->43->60 then removeDuplicates() should convert the list to 11->21->43->60.

Algorithm:
Traverse the list from the head (or start) node. While traversing, compare each node with its next node. If the data of the next node is the same as the current node then delete the next node. Before we delete a node, we need to store the next pointer of the node

Implementation:
Functions other than removeDuplicates() are just to create a linked list and test removeDuplicates().

Python3

# Python3 program to remove duplicate
# nodes from a sorted linked list 
 
# Node class 
class Node: 
 
    # Constructor to initialize 
    # the node object 
    def __init__(self, data): 
        self.data = data 
        self.next = None
 
class LinkedList: 
 
    # Function to initialize head 
    def __init__(self): 
        self.head = None
 
    # Function to insert a new node 
    # at the beginning 
    def push(self, new_data): 
        new_node = Node(new_data) 
        new_node.next = self.head 
        self.head = new_node 
 
    # Given a reference to the head of a 
    # list and a key, delete the first 
    # occurrence of key in linked list 
    def deleteNode(self, key): 
         
        # Store head node 
        temp = self.head 
 
        # If head node itself holds the 
        # key to be deleted 
        if (temp is not None): 
            if (temp.data == key): 
                self.head = temp.next
                temp = None
                return
 
        # Search for the key to be deleted, 
        # keep track of the previous node as
        # we need to change 'prev.next' 
        while(temp is not None): 
            if temp.data == key: 
                break
            prev = temp 
            temp = temp.next
 
        # If key was not present in 
        # linked list 
        if(temp == None): 
            return
 
        # Unlink the node from linked list 
        prev.next = temp.next
 
        temp = None
 
    # Utility function to print the 
    # linked LinkedList 
    def printList(self): 
        temp = self.head 
        while(temp): 
            print(temp.data , end = ' ')
            temp = temp.next
     
    # This function removes duplicates 
    # from a sorted list         
    def removeDuplicates(self):
        temp = self.head
        if temp is None:
            return
        while temp.next is not None:
            if temp.data == temp.next.data:
                new = temp.next.next
                temp.next = None
                temp.next = new
            else:
                temp = temp.next
        return self.head
 
# Driver Code 
llist = LinkedList() 
 
llist.push(20) 
llist.push(13) 
llist.push(13)
llist.push(11)
llist.push(11)
llist.push(11)
print ("Created Linked List: ")
llist.printList() 
print()
print("Linked List after removing", 
      "duplicate elements:")
llist.removeDuplicates()
llist.printList() 
# This code is contributed by Dushyant Pathak.

Output:

Linked list before duplicate removal  11 11 11 13 13 20
Linked list after duplicate removal  11 13 20

Time Complexity: O(n) where n is the number of nodes in the given linked list.

Auxiliary Space: O(1)

Recursive Approach :

Python3

# Python3 Program to remove duplicates
# from a sorted linked list 
import math
 
# Link list node 
class Node: 
    def __init__(self, data): 
        self.data = data 
        self.next = None
 
# The function removes duplicates 
# from a sorted list 
def removeDuplicates(head): 
     
    # Pointer to store the pointer 
    # of a node to be deleted to_free 
     
    # Do nothing if the list is empty 
    if (head == None): 
        return
 
    # Traverse the list till last node 
    if (head.next != None): 
         
        # Compare head node with next node 
        if (head.data == head.next.data): 
             
            # The sequence of steps is important. 
            # to_free pointer stores the next of head 
            # pointer which is to be deleted. 
            to_free = head.next
            head.next = head.next.next
             
            # free(to_free) 
            removeDuplicates(head) 
         
        # This is tricky: only advance if no deletion
        else: 
            removeDuplicates(head.next) 
         
    return head
 
# UTILITY FUNCTIONS 
# Function to insert a node at the 
# beginning of the linked list 
def push(head_ref, new_data): 
     
    # Allocate node 
    new_node = Node(new_data) 
             
    # Put in the data 
    new_node.data = new_data 
                 
    # Link the old list off the 
    # new node 
    new_node.next = head_ref     
         
    # Move the head to point to the 
    # new node 
    head_ref = new_node
    return head_ref
 
# Function to print nodes in a given 
# linked list 
def printList(node): 
    while (node != None): 
        print(node.data, end = " ") 
        node = node.next
     
# Driver code
if __name__=='__main__': 
 
    # Start with the empty list 
    head = None
     
    # Let us create a sorted linked list
    # to test the functions 
    # Created linked list will be 
    # 11.11.11.13.13.20 
    head = push(head, 20) 
    head = push(head, 13) 
    head = push(head, 13) 
    head = push(head, 11) 
    head = push(head, 11) 
    head = push(head, 11)                                 
 
    print("Linked list before duplicate removal ",
           end = "") 
    printList(head) 
 
    # Remove duplicates from linked list 
    removeDuplicates(head) 
 
    print("Linked list after duplicate removal ",
           end = "") 
    printList(head)         
     
# This code is contributed by Srathore

Output:

Linked list before duplicate removal  11 11 11 13 13 20
Linked list after duplicate removal  11 13 20

Time Complexity: O(n), where n is the number of nodes in the given linked list.

Auxiliary Space: O(n), due to recursive stack where n is the number of nodes in the given linked list.

Another Approach: Create a pointer that will point towards the first occurrence of every element and another pointer temp which will iterate to every element and when the value of the previous pointer is not equal to the temp pointer, we will set the pointer of the previous pointer to the first occurrence of another node.

Below is the implementation of the above approach:

Python3

# Python3 program to remove duplicates 
# from a sorted linked list  
import math 
   
# Link list node  
class Node:  
     
    def __init__(self, data):        
        self.data = data  
        self.next = None
   
# The function removes duplicates  
# from the given linked list 
def removeDuplicates(head):  
     
    # Do nothing if the list consist of 
    # only one element or empty  
    if (head == None and
        head.next == None):
        return
     
    # Construct a pointer 
    # pointing towards head
    current = head
     
    # Initialise a while loop till the 
    # second last node of the linkedlist 
    while (current.next):
         
        # If the data of current and next
        # node is equal we will skip the
        # node between them
        if current.data == current.next.data:
            current.next = current.next.next
             
        # If the data of current and 
        # next node is different move
        # the pointer to the next node
        else:
            current = current.next
     
    return
   
# UTILITY FUNCTIONS  
# Function to insert a node at the  
# beginning of the linked list  
def push(head_ref, new_data):  
     
    # Allocate node  
    new_node = Node(new_data)  
               
    # Put in the data  
    new_node.data = new_data  
                   
    # Link the old list off 
    # the new node  
    new_node.next = head_ref      
           
    # Move the head to point
    # to the new node  
    head_ref = new_node 
     
    return head_ref 
   
# Function to print nodes 
# in a given linked list  
def printList(node):  
     
    while (node != None):  
        print(node.data, end = " ")  
        node = node.next
       
# Driver code 
if __name__=='__main__':  
   
    head = None   
    head = push(head, 20)  
    head = push(head, 13)  
    head = push(head, 13)  
    head = push(head, 11)  
    head = push(head, 11)  
    head = push(head, 11)                                  
   
    print("List before removal of " 
          "duplicates ", end = "")  
    printList(head)  
   
    removeDuplicates(head)  
   
    print("List after removal of " 
          "elements ", end = "")            
    printList(head)          
# This code is contributed by MukulTomar

Output:

List before removal of duplicates
11 11 11 13 13 20 
List after removal of elements
11 13 20

Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Remove duplicates from a sorted linked list for more details!