Remove duplicates from a sorted linked list

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(). 

C++

/* C++ Program to remove duplicates from a sorted linked
 * list */
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
class Node {
public:
    int data;
    Node* next;
};

/* The function removes duplicates from a sorted list */
void removeDuplicates(Node* head)
{
    /* Pointer to traverse the linked list */
    Node* current = head;

    /* Pointer to store the next pointer of a node to be
     * deleted*/
    Node* next_next;

    /* do nothing if the list is empty */
    if (current == NULL)
        return;

    /* Traverse the list till last node */
    while (current->next != NULL) {
        /* Compare current node with next node */
        if (current->data == current->next->data) {
            /* The sequence of steps is important*/
            next_next = current->next->next;
            free(current->next);
            current->next = next_next;
        }
        else /* This is tricky: only advance if no deletion
              */
        {
            current = current->next;
        }
    }
}

/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning of the linked
 * list */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = new Node();

    /* put in the data */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

/* Function to print nodes in a given linked list */
void printList(Node* node)
{
    while (node != NULL) {
        cout << " " << node->data;
        node = node->next;
    }
}

/* Driver program to test above functions*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;

    /* Let us create a sorted linked list to test the
    functions Created linked list will be
    11->11->11->13->13->20 */
    push(&head, 20);
    push(&head, 13);
    push(&head, 13);
    push(&head, 11);
    push(&head, 11);
    push(&head, 11);

    cout << "Linked list before duplicate removal " << endl;
    printList(head);

    /* Remove duplicates from linked list */
    removeDuplicates(head);

    cout << "\nLinked list after duplicate removal "
         << endl;
    printList(head);

    return 0;
}

// This code is contributed by rathbhupendra

C

/* C Program to remove duplicates from a sorted linked list
 */
#include <stdio.h>
#include <stdlib.h>

/* Link list node */
struct Node {
    int data;
    struct Node* next;
};

/* The function removes duplicates from a sorted list */
void removeDuplicates(struct Node* head)
{
    /* Pointer to traverse the linked list */
    struct Node* current = head;

    /* Pointer to store the next pointer of a node to be
     * deleted*/
    struct Node* next_next;

    /* do nothing if the list is empty */
    if (current == NULL)
        return;

    /* Traverse the list till last node */
    while (current->next != NULL) {
        /* Compare current node with next node */
        if (current->data == current->next->data) {
            /* The sequence of steps is important*/
            next_next = current->next->next;
            free(current->next);
            current->next = next_next;
        }
        else /* This is tricky: only advance if no deletion
              */
        {
            current = current->next;
        }
    }
}

/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning of the linked
 * list */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));

    /* put in the data  */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

/* Function to print nodes in a given linked list */
void printList(struct Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}

/* Driver program to test above functions*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;

    /* Let us create a sorted linked list to test the
     functions Created linked list will be
     11->11->11->13->13->20 */
    push(&head, 20);
    push(&head, 13);
    push(&head, 13);
    push(&head, 11);
    push(&head, 11);
    push(&head, 11);

    printf("\n Linked list before duplicate removal \n");
    printList(head);

    /* Remove duplicates from linked list */
    removeDuplicates(head);

    printf("\n Linked list after duplicate removal \n");
    printList(head);

    return 0;
}

Java

// Java program to remove duplicates from a sorted linked
// list
import java.io.*;
class LinkedList {
    Node head; // head of list

    /* Linked list Node*/
    class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }

    void removeDuplicates()
    {
        /*Another reference to head*/
        Node curr = head;

        /* Traverse list till the last node */
        while (curr != null) {
            Node temp = curr;
            /*Compare current node with the next node and
            keep on deleting them until it matches the
            current node data */
            while (temp != null && temp.data == curr.data) {
                temp = temp.next;
            }
            /*Set current node next to the next different
            element denoted by temp*/
            curr.next = temp;
            curr = curr.next;
        }
    }

    /* Utility functions */

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Function to print linked list */
    void printList()
    {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }

    /* Driver program to test above functions */
    public static void main(String args[])
    {
        LinkedList llist = new LinkedList();
        llist.push(20);
        llist.push(13);
        llist.push(13);
        llist.push(11);
        llist.push(11);
        llist.push(11);

        System.out.println(
            "List before removal of duplicates");
        llist.printList();

        llist.removeDuplicates();

        System.out.println(
            "List after removal of elements");
        llist.printList();
    }
}
/* This code is contributed by Rajat Mishra */

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.

C#

// C# program to remove duplicates
// from a sorted linked list
using System;

public class LinkedList {
    Node head; // head of list

    /* Linked list Node*/
    class Node {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    void removeDuplicates()
    {
        /*Another reference to head*/
        Node current = head;

        /* Pointer to store the next
        pointer of a node to be deleted*/
        Node next_next;

        /* do nothing if the list is empty */
        if (head == null)
            return;

        /* Traverse list till the last node */
        while (current.next != null) {

            /*Compare current node with the next node */
            if (current.data == current.next.data) {
                next_next = current.next.next;
                current.next = null;
                current.next = next_next;
            }
            else // advance if no deletion
                current = current.next;
        }
    }

    /* Utility functions */

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Function to print linked list */
    void printList()
    {
        Node temp = head;
        while (temp != null) {
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
        Console.WriteLine();
    }

    /* Driver code */
    public static void Main(String[] args)
    {
        LinkedList llist = new LinkedList();
        llist.push(20);
        llist.push(13);
        llist.push(13);
        llist.push(11);
        llist.push(11);
        llist.push(11);

        Console.WriteLine(
            "List before removal of duplicates");
        llist.printList();

        llist.removeDuplicates();

        Console.WriteLine("List after removal of elements");
        llist.printList();
    }
}

/* This code is contributed by 29AjayKumar */

Javascript

<script>
// Javascript program to remove duplicates from a sorted linked list

     /* Linked list Node*/
    class Node
    {
        constructor(d) {
            this.data = d;
            this.next = null;
          }
    }

    let head=new Node();    // head of list
    function removeDuplicates()
    {
        /*Another reference to head*/
        let curr = head;

        /* Traverse list till the last node */
        while (curr != null) {
             let temp = curr;
            /*Compare current node with the next node and
            keep on deleting them until it matches the current
            node data */
            while(temp!=null && temp.data==curr.data) {
                temp = temp.next;
            }
            /*Set current node next to the next different
            element denoted by temp*/
            curr.next = temp;
            curr = curr.next;
        }
    }

    /* Utility functions */

    /* Inserts a new Node at front of the list. */
    function push(new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        let new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Function to print linked list */
    function printList()
    {
        let temp = head;
         while (temp != null && temp.data)
         {

            document.write(temp.data+" ");
            temp = temp.next;
         }
         document.write("<br>");
    }

    /* Driver program to test above functions */
    push(20)
    push(13)
    push(13)
    push(11)
    push(11)
    push(11)

    document.write("List before removal of duplicates ");
    printList();

    removeDuplicates();

    document.write("List after removal of elements ");
    printList();


    // This code is contributed by unknown2108
</script>

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) , as there is no extra space used.

Recursive Approach :

C++

/* C++ Program to remove duplicates
from a sorted linked list */
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
class Node {
public:
    int data;
    Node* next;
};

/* The function removes duplicates
from a sorted list */
void removeDuplicates(Node* head)
{
    /* Pointer to store the pointer of a node to be
     * deleted*/
    Node* to_free;

    /* do nothing if the list is empty */
    if (head == NULL)
        return;

    /* Traverse the list till last node */
    if (head->next != NULL) {

        /* 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);
        }
        else /* This is tricky: only
        advance if no deletion */
        {
            removeDuplicates(head->next);
        }
    }
}

/* UTILITY FUNCTIONS */
/* Function to insert a node at the
beginning of the linked list */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = new Node();

    /* put in the data */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

/* Function to print nodes
in a given linked list */
void printList(Node* node)
{
    while (node != NULL) {
        cout << " " << node->data;
        node = node->next;
    }
}

/* Driver code*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;

    /* Let us create a sorted linked
    list to test the functions
    Created linked list will be
    11->11->11->13->13->20 */
    push(&head, 20);
    push(&head, 13);
    push(&head, 13);
    push(&head, 11);
    push(&head, 11);
    push(&head, 11);

    cout << "Linked list before duplicate removal ";
    printList(head);

    /* Remove duplicates from linked list */
    removeDuplicates(head);

    cout << "\nLinked list after duplicate removal ";
    printList(head);

    return 0;
}
// This code is contributed by Ashita Gupta

C

/* C recursive Program to remove duplicates from a sorted
 * linked list */
#include <stdio.h>
#include <stdlib.h>

/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning of the linked
 * list */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));

    /* put in the data  */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

/* Function to print nodes in a given linked list */
void printList(struct Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}

Node* deleteDuplicates(Node* head)
{
    if (head == nullptr)
        return nullptr;
    if (head->next == nullptr)
        return head;

    if (head->data == head->next->data) {
        Node* tmp;
        // If find next element duplicate, preserve the next
        // pointer to be deleted, skip it, and then delete
        // the stored one. Return head
        tmp = head->next;
        head->next = head->next->next;
        free(tmp);
        return deleteDuplicates(head);
    }

    else {

        // if doesn't find next element duplicate, leave
        // head and check from next element
        head->next = deleteDuplicates(head->next);
        return head;
    }
}

int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;

    /* Let us create a sorted linked list to test the
     functions Created linked list will be
     11->11->11->13->13->20 */
    push(&head, 20);
    push(&head, 13);
    push(&head, 13);
    push(&head, 11);
    push(&head, 11);
    push(&head, 11);

    printf("\n Linked list before duplicate removal  ");
    printList(head);

    /* Remove duplicates from linked list */
    head = deleteDuplicates(head);

    printf("\n Linked list after duplicate removal ");
    printList(head);

    return 0;
}
/* This code is contributed by Yogesh shukla */

Java

// Java Program to remove duplicates
// from a sorted linked list
import java.io.*;
class GFG {
    /* Link list node */
    static class Node {
        int data;
        Node next;
    };

    // The function removes duplicates
    // from a sorted list
    static Node removeDuplicates(Node head)
    {
        /* Pointer to store the pointer
        of a node to be deleted*/
        Node to_free;

        /* do nothing if the list is empty */
        if (head == null)
            return null;

        /* Traverse the list till last node */
        if (head.next != null) {

            /* 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;
                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 */
    static Node push(Node head_ref, int new_data)
    {
        /* allocate node */
        Node new_node = new Node();

        /* put in the data */
        new_node.data = new_data;

        /* link the old list of 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 */
    static void printList(Node node)
    {
        while (node != null) {
            System.out.print(" " + node.data);
            node = node.next;
        }
    }

    /* Driver code*/
    public static void main(String args[])
    {
        /* Start with the empty list */
        Node head = null;

        /* 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);

        System.out.println("Linked list before"
                           + " duplicate removal ");
        printList(head);

        /* Remove duplicates from linked list */
        head = removeDuplicates(head);

        System.out.println("\nLinked list after"
                           + " duplicate removal ");
        printList(head);
    }
}

// This code is contributed by Arnab Kundu

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 of 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("\nLinked list after duplicate removal ",
          end="")
    printList(head)

# This code is contributed by Srathore

C#

// C# Program to remove duplicates
// from a sorted linked list
using System;

class GFG {
    /* Link list node */
    public class Node {
        public int data;
        public Node next;
    };

    // The function removes duplicates
    // from a sorted list
    static Node removeDuplicates(Node head)
    {
        /* Pointer to store the pointer
        of a node to be deleted*/
        Node to_free;

        /* do nothing if the list is empty */
        if (head == null)
            return null;

        /* Traverse the list till last node */
        if (head.next != null) {

            /* 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;
                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 */
    static Node push(Node head_ref, int new_data)
    {
        /* allocate node */
        Node new_node = new Node();

        /* put in the data */
        new_node.data = new_data;

        /* link the old list of 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 */
    static void printList(Node node)
    {
        while (node != null) {
            Console.Write(" " + node.data);
            node = node.next;
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        /* Start with the empty list */
        Node head = null;

        /* 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);

        Console.Write("Linked list before"
                      + " duplicate removal ");
        printList(head);

        /* Remove duplicates from linked list */
        head = removeDuplicates(head);

        Console.Write("\nLinked list after"
                      + " duplicate removal ");
        printList(head);
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript Program to remove duplicates
// from a sorted linked list

    /* Link list node */
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
    // The function removes duplicates
    // from a sorted list
    function removeDuplicates(head) {
        /*
         Pointer to store the pointer
         of a node to be deleted
         */
       var to_free;

        /* do nothing if the list is empty */
        if (head == null)
            return null;

        /* Traverse the list till last node */
        if (head.next != null) {

            /* 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;
                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
     */
    function push(head_ref , new_data) {
        /* allocate node */
var new_node = new Node();

        /* put in the data */
        new_node.data = new_data;

        /* link the old list of 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 */
    function printList(node) {
        while (node != null) {
            document.write(" " + node.data);
            node = node.next;
        }
    }

    /* Driver code */

        /* Start with the empty list */
var head = null;

        /*
          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);

        document.write("Linked list before" +
        " duplicate removal <br/>");
        printList(head);

        /* Remove duplicates from linked list */
        head = removeDuplicates(head);

        document.write("<br/>Linked list after" +
        " duplicate removal <br/>");
        printList(head);

// This code is contributed by todaysgaurav

</script>

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)

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:

C++14

// C++ program to remove duplicates
// from a sorted linked list
#include <bits/stdc++.h>
using namespace std;

// Linked list Node
struct Node {
    int data;
    Node* next;
    Node(int d)
    {
        data = d;
        next = NULL;
    }
};

// Function to remove duplicates
// from the given linked list
Node* removeDuplicates(Node* head)
{
    // Two references to head temp will iterate to the whole
    // Linked List prev will point towards the first
    // occurrence of every element
    Node *temp = head, *prev = head;
    // Traverse list till the last node
    while (temp != NULL) {
        // Compare values of both pointers
        if (temp->data != prev->data) {
            // if the value of prev is not equal to the
            // value of temp that means there are no more
            // occurrences of the prev data-> So we can set
            // the next of prev to the temp node->*/
            prev->next = temp;
            prev = temp;
        }
        // Set the temp to the next node
        temp = temp->next;
    }
    // This is the edge case if there are more than one
    // occurrences of the last element
    if (prev != temp)
        prev->next = NULL;
    return head;
}

Node* push(Node* head, int new_data)
{
    /* 1 & 2: Allocate the Node & Put in the data*/
    Node* new_node = new Node(new_data);
    /* 3. Make next of new Node as head */
    new_node->next = head;
    /* 4. Move the head to point to new Node */
    head = new_node;
    return head;
}

/* Function to print linked list */
void printList(Node* head)
{
    Node* temp = head;
    while (temp != NULL) {
        cout << temp->data << " ";
        temp = temp->next;
    }
    cout << endl;
}

/* Driver code */
int main()
{
    Node* llist = NULL;
    llist = push(llist, 20);
    llist = push(llist, 13);
    llist = push(llist, 13);
    llist = push(llist, 11);
    llist = push(llist, 11);
    llist = push(llist, 11);
    cout << ("List before removal of duplicates\n");
    printList(llist);
    cout << ("List after removal of elements\n");
    llist = removeDuplicates(llist);
    printList(llist);
}

// This code is contributed by Sania Kumari Gupta

C

// C program to remove duplicates from a sorted linked list
#include <stdio.h>
#include <stdlib.h>

/* Link list node */
typedef struct Node {
    int data;
    struct Node* next;
} Node;

// Function to remove duplicates
// from the given linked list
Node* removeDuplicates(Node* head)
{
    // Two references to head temp will iterate to the whole
    // Linked List prev will point towards the first
    // occurrence of every element
    Node *temp = head, *prev = head;
    // Traverse list till the last node
    while (temp != NULL) {
        // Compare values of both pointers
        if (temp->data != prev->data) {
            // if the value of prev is not equal to the
            // value of temp that means there are no more
            // occurrences of the prev data-> So we can set
            // the next of prev to the temp node->*/
            prev->next = temp;
            prev = temp;
        }
        // Set the temp to the next node
        temp = temp->next;
    }
    // This is the edge case if there are more than one
    // occurrences of the last element
    if (prev != temp)
        prev->next = NULL;
    return head;
}

void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = (Node*)malloc(sizeof(Node));
    /* put in the data  */
    new_node->data = new_data;
    /* link the old list of the new node */
    new_node->next = (*head_ref);
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

/* Function to print linked list */
void printList(Node* head)
{
    Node* temp = head;
    while (temp != NULL) {
        printf("%d ", temp->data);
        temp = temp->next;
    }
    printf("\n");
}

/* Driver code */
int main()
{
    Node* llist = NULL;
    push(&llist, 20);
    push(&llist, 13);
    push(&llist, 13);
    push(&llist, 11);
    push(&llist, 11);
    push(&llist, 11);
    printf("List before removal of duplicates\n");
    printList(llist);
    printf("List after removal of elements\n");
    llist = removeDuplicates(llist);
    printList(llist);
}

// This code is contributed by Sania Kumari Gupta

Java

// Java program to remove duplicates
// from a sorted linked list
import java.io.*;
class LinkedList {
    // head of list
    Node head;
    // Linked list Node
    class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }

    // Function to remove duplicates from the given linked
    // list
    void removeDuplicates()
    {
        // Two references to head temp will iterate to the
        // whole Linked List prev will point towards the
        // first occurrence of every element
        Node temp = head, prev = head;

        // Traverse list till the last node
        while (temp != null) {

            // Compare values of both pointers
            if (temp.data != prev.data) {
                // if the value of prev is not equal to the
                // value of temp that means there are no
                // more occurrences of the prev data. So we
                // can set the next of prev to the temp
                // node.
                prev.next = temp;
                prev = temp;
            }
            // Set the temp to the next node
            temp = temp.next;
        }
        // This is the edge case if there are more than one
        // occurrences of the last element
        if (prev != temp)
            prev.next = null;
    }

    /* Utility functions */

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);
        /* 3. Make next of new Node as head */
        new_node.next = head;
        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Function to print linked list */
    void printList()
    {
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }

    /* Driver program to test above functions */
    public static void main(String args[])
    {
        LinkedList llist = new LinkedList();
        llist.push(20);
        llist.push(13);
        llist.push(13);
        llist.push(11);
        llist.push(11);
        llist.push(11);

        System.out.print("List before ");
        System.out.println("removal of duplicates");
        llist.printList();

        llist.removeDuplicates();

        System.out.println(
            "List after removal of elements");
        llist.printList();
    }
}

// This code is contributed by Sania Kumari Gupta

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 of
    # 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("\nList after removal of "
          "elements ", end="")

    printList(head)

# This code is contributed by MukulTomar

C#

// C# program to remove duplicates
// from a sorted linked list
using System;

class LinkedList {

    // head of list
    public Node head;

    // Linked list Node
    public class Node {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    // Function to remove duplicates
    // from the given linked list
    void removeDuplicates()
    {

        // Two references to head
        // temp will iterate to the
        // whole Linked List
        // prev will point towards
        // the first occurrence of every element
        Node temp = head, prev = head;

        // Traverse list till the last node
        while (temp != null) {

            // Compare values of both pointers
            if (temp.data != prev.data) {

                /* if the value of prev is
                not equal to the value of
                temp that means there are no
                more occurrences of the prev data.
                So we can set the next of
                prev to the temp node.*/
                prev.next = temp;
                prev = temp;
            }

            /*Set the temp to the next node*/
            temp = temp.next;
        }

        /*This is the edge case if there
        are more than one occurrences
        of the last element*/
        if (prev != temp) {
            prev.next = null;
        }
    }

    /* Utility functions */

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Function to print linked list */
    void printList()
    {
        Node temp = head;
        while (temp != null) {
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
        Console.WriteLine();
    }

    /* Driver code */
    public static void Main(string[] args)
    {
        LinkedList llist = new LinkedList();
        llist.push(20);
        llist.push(13);
        llist.push(13);
        llist.push(11);
        llist.push(11);
        llist.push(11);
        Console.Write("List before ");
        Console.WriteLine("removal of duplicates");
        llist.printList();
        llist.removeDuplicates();
        Console.WriteLine("List after removal of elements");
        llist.printList();
    }
}

// This code is contributed by rutvik_56

Javascript

<script>
// javascript program to remove duplicates
// from a sorted linked list

    // head of list
    var head;

    // Linked list Node
     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }


    // Function to remove duplicates
    // from the given linked list
    function removeDuplicates() {
        // Two references to head
        // temp will iterate to the
        // whole Linked List
        // prev will point towards
        // the first occurrence of every element
        var temp = head, prev = head;

        // Traverse list till the last node
        while (temp != null) {

            // Compare values of both pointers
            if (temp.data != prev.data) {
                /*
                 * if the value of prev is not equal to the value of temp that means there are
                 * no more occurrences of the prev data. So we can set the next of prev to the
                 * temp node.
                 */
                prev.next = temp;
                prev = temp;
            }
            /* Set the temp to the next node */
            temp = temp.next;
        }
        /*
         * This is the edge case if there are more than one occurrences of the last
         * element
         */
        if (prev != temp) {
            prev.next = null;
        }
    }

    /* Utility functions */

    /* Inserts a new Node at front of the list. */
    function push(new_data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
        var new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Function to print linked list */
    function printList() {
        var temp = head;
        while (temp != null) {
            document.write(temp.data + " ");
            temp = temp.next;
        }
        document.write("<br/>");
    }

    /* Driver program to test above functions */

        push(20);
        push(13);
        push(13);
        push(11);
        push(11);
        push(11);

        document.write("List before ");
        document.write("removal of duplicates<br/>");
        printList();

        removeDuplicates();

        document.write("List after removal of elements<br/>");
        printList();

// This code contributed by aashish1995
</script>

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) 

Another Approach: Using Maps

The idea is to push all the values in a map and printing its keys.

Below is the implementation of the above approach:

C++

// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
struct Node {
    int data;
    Node* next;
    Node()
    {
        data = 0;
        next = NULL;
    }
};

/* Function to insert a node at
   the beginning of the linked
 * list */
void push(Node** head_ref, int new_data)
{

    /* allocate node */
    Node* new_node = new Node();

    /* put in the data */
    new_node->data = new_data;

    /* link the old list of
    the new node */
    new_node->next = (*head_ref);

    /* move the head to point
    to the new node */
    (*head_ref) = new_node;
}

/* Function to print nodes
in a given linked list */
void printList(Node* node)
{
    while (node != NULL) {
        cout << node->data << " ";
        node = node->next;
    }
}

// Function to remove duplicates
void removeDuplicates(Node* head)
{
    unordered_map<int, bool> track;
    Node* temp = head;
    while (temp) {
        if (track.find(temp->data) == track.end()) {
            cout << temp->data << " ";
        }
        track[temp->data] = true;
        temp = temp->next;
    }
}

// Driver Code
int main()
{
    Node* head = NULL;

    /* Created linked list will be
    11->11->11->13->13->20 */
    push(&head, 20);
    push(&head, 13);
    push(&head, 13);
    push(&head, 11);
    push(&head, 11);
    push(&head, 11);

    cout << "Linked list before duplicate removal ";
    printList(head);

    cout << "\nLinked list after duplicate removal ";
    removeDuplicates(head);

    return 0;
}

// This code is contributed by yashbeersingh42

Java

// Java program for the above approach
import java.io.*;
import java.util.*;

class Node {
    int data;
    Node next;
    Node()
    {
        data = 0;
        next = null;
    }
}
class GFG {

    /* Function to insert a node at
   the beginning of the linked
 * list */
    static Node push(Node head_ref, int new_data)
    {

        /* allocate node */
        Node new_node = new Node();

        /* put in the data */
        new_node.data = new_data;

        /* link the old list of
        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 */
    static void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    // Function to remove duplicates
    static void removeDuplicates(Node head)
    {
        HashMap<Integer, Boolean> track = new HashMap<>();
        Node temp = head;

        while (temp != null) {
            if (!track.containsKey(temp.data)) {
                System.out.print(temp.data + " ");
            }
            track.put(temp.data, true);
            temp = temp.next;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        Node head = null;

        /* 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);
        System.out.print(
            "Linked list before duplicate removal ");
        printList(head);
        System.out.print(
            "\nLinked list after duplicate removal  ");
        removeDuplicates(head);
    }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python program for the above approach
class Node:
    def __init__(self):
        self.data = 0
        self.next = None

# Function to insert a node at
# the beginning of the linked list


def push(head_ref, new_data):

    # allocate node
    new_node = Node()

    # put in the data
    new_node.data = new_data

    # link the old list of 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


# Function to remove duplicates
def removeDuplicates(head):
    track = {}
    temp = head

    while(temp != None):
        if (not temp.data in track):
            print(temp.data, end=" ")

        track[temp.data] = True
        temp = temp.next


# Driver Code
head = None

# 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)
print("\nLinked list after duplicate removal  ", end=" ")
removeDuplicates(head)

# This code is contributed by _Saurabh_jaiswal

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

public class Node {
    public int data;
    public Node next;
    public Node()
    {
        data = 0;
        next = null;
    }
}
public class GFG {

    /* Function to insert a node at
   the beginning of the linked
 * list */
    static Node push(Node head_ref, int new_data)
    {

        /* allocate node */
        Node new_node = new Node();

        /* put in the data */
        new_node.data = new_data;

        /* link the old list of
        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 */
    static void printList(Node node)
    {
        while (node != null) {
            Console.Write(node.data + " ");
            node = node.next;
        }
    }

    // Function to remove duplicates
    static void removeDuplicates(Node head)
    {
        Dictionary<int, bool> track
            = new Dictionary<int, bool>();
        Node temp = head;
        while (temp != null) {
            if (!track.ContainsKey(temp.data)) {
                Console.Write(temp.data + " ");
                track.Add(temp.data, true);
            }

            temp = temp.next;
        }
    }

    // Driver Code
    static public void Main()
    {
        Node head = null;

        /* 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);

        Console.Write(
            "Linked list before duplicate removal ");
        printList(head);
        Console.Write(
            "\nLinked list after duplicate removal  ");
        removeDuplicates(head);
    }
}

// This code is contributed by rag2127

Javascript

<script>

// Javascript program for the above approach
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
}

/* Function to insert a node at
   the beginning of the linked
 * list */
function push(head_ref, new_data)
{

    /* allocate node */
    let new_node = new Node();

    /* put in the data */
    new_node.data = new_data;

    /* link the old list of
    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 */
function printList(node)
{
    while (node != null)
    {
        document.write(node.data + " ");
        node = node.next;
    }
}

// Function to remove duplicates
function removeDuplicates(head)
{
    let track = new Map();
    let temp = head;

    while(temp != null)
    {
        if (!track.has(temp.data))
        {
            document.write(temp.data + " ");
        }
        track.set(temp.data, true);
        temp = temp.next;
    }
}

// Driver Code
let head = null;

/* 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);
document.write("Linked list before duplicate removal ");
printList(head);
document.write("<br>Linked list after duplicate removal  ");
removeDuplicates(head);

// This code is contributed by patel2127

</script>

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) 

Using Set data structure:

We know that set data structure always stores the unique element.

so, the intuition comes from there-

Explanation:

The code then defines a class Solution which contains the deleteDuplicates function. The function takes a pointer to the head of the linked list as an argument. If the head is empty, it returns it. Otherwise, it initializes a set data structure to store unique values in the linked list. It also declares two pointers curr and prev to traverse the linked list and keep track of the previous node, respectively.

The code then enters a while loop that continues as long as curr is not NULL. In each iteration, it checks if the value of the current node already exists in the set. If it does, the code updates the next pointer of the previous node to skip the current node and remove it from the linked list. If the value does not exist in the set, it is inserted into the set and prev is updated to become the current node. After each iteration, curr is updated to point to the next node.

Finally, the code returns the head of the linked list.

The main function demonstrates how to use the deleteDuplicates function by creating a linked list with values 1, 2, 2,2, 3 ,3,3and calling the function to remove duplicates. It then prints the linked list after duplicates are removed.

Note: The code assumes that the linked list is sorted and the duplicates appear consecutively

Below is the implementation for above approach : 

C++

#include <iostream>
#include <set>

// Define a node in a linked list
struct ListNode {
    int val; // Value of the node
    ListNode*
        next; // Pointer to the next node in the linked list
    ListNode(int x)
    {
        this->val = x;
        this->next = NULL;
    }
};

class Solution {
public:
    // Remove duplicates from sorted linked list using set
    ListNode* deleteDuplicates(ListNode* head)
    {
        // Return head if it's empty
        if (!head)
            return head;

        // Use set to store unique values in linked list
        std::set<int> set;
        ListNode* curr
            = head; // Pointer to traverse the linked list
        ListNode* prev = NULL; // Pointer to keep track of
                               // the previous node

        // Iterate through the linked list
        while (curr) {
            // If the current value already exists in set,
            // remove the node
            if (set.count(curr->val)) {
                prev->next = curr->next;
            }
            else {
                // Otherwise, add the value to set and move
                // on to the next node
                set.insert(curr->val);
                prev = curr;
            }
            curr = curr->next;
        }

        // Return the head of the linked list
        return head;
    }
};

int main()
{
    // Initialize linked list with values 1, 2, 2, 3
    ListNode* head = new ListNode(1);
    head->next = new ListNode(2);
    head->next->next = new ListNode(2);
    head->next->next->next = new ListNode(2);
    head->next->next->next->next = new ListNode(3);
    head->next->next->next->next->next = new ListNode(3);
    head->next->next->next->next->next->next
        = new ListNode(3);
    ListNode* BeforePrinter = head;
    std::cout
        << "Before removing duplicates linked list is:"
        << std::endl;
    while (BeforePrinter) {
        std::cout << BeforePrinter->val << " ";
        BeforePrinter = BeforePrinter->next;
    }
    std::cout << std::endl;

    Solution solution;
    // Remove duplicates from the linked list using the
    // deleteDuplicates function
    head = solution.deleteDuplicates(head);

    // Print the linked list after removing duplicates
    std::cout << "after removing duplicates linked list is:"
              << std::endl;
    ListNode* curr = head;
    while (curr) {
        std::cout << curr->val << " ";
        curr = curr->next;
    }
    std::cout << std::endl;

    return 0;
}
// This code is contributed by Veerendra Singh Rajpoot

Python3

# Define a node in a linked list
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        # Return head if it's empty
        if not head:
            return head

        # Use set to store unique values in linked list
        unique_vals = set()
        curr = head  # Pointer to traverse the linked list
        prev = None  # Pointer to keep track of the previous node

        # Iterate through the linked list
        while curr:
            # If the current value already exists in set, remove the node
            if curr.val in unique_vals:
                prev.next = curr.next
            else:
                # Otherwise, add the value to set and move on to the next node
                unique_vals.add(curr.val)
                prev = curr
            curr = curr.next

        # Return the head of the linked list
        return head


# Initialize linked list with values 1, 2, 2, 3
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(2)
head.next.next.next = ListNode(2)
head.next.next.next.next = ListNode(3)
head.next.next.next.next.next = ListNode(3)
head.next.next.next.next.next.next = ListNode(3)

BeforePrinter = head
print("Before removing duplicates linked list is:")
while BeforePrinter:
    print(BeforePrinter.val, end=" ")
    BeforePrinter = BeforePrinter.next
print()

solution = Solution()
# Remove duplicates from the linked list using the deleteDuplicates function
head = solution.deleteDuplicates(head)

# Print the linked list after removing duplicates
print("After removing duplicates linked list is:")
curr = head
while curr:
    print(curr.val, end=" ")
    curr = curr.next
print()

C#

using System;
using System.Collections.Generic;

// Define a node in a linked list
class ListNode {
    // Value of the node
    public int val;
    // Pointer to the next node in the linked list
    public ListNode next;
    public ListNode(int x)
    {
        this.val = x;
        this.next = null;
    }
}

class Solution {
    // Remove duplicates from sorted linked list using set
    public ListNode DeleteDuplicates(ListNode head)
    {
        // Return head if it's empty
        if (head == null)
            return head;

        // Use set to store unique values in linked list
        HashSet<int> set = new HashSet<int>();
        // Pointer to traverse the linked list
        ListNode curr = head;
        // Pointer to keep track of the previous node
        ListNode prev = null;

        // Iterate through the linked list
        while (curr != null) {
            // If the current value already exists in set,
            // remove the node
            if (set.Contains(curr.val)) {
                prev.next = curr.next;
            }
            else {
                // Otherwise, add the value to set and move
                // on to the next node
                set.Add(curr.val);
                prev = curr;
            }
            curr = curr.next;
        }

        // Return the head of the linked list
        return head;
    }
}

class Program {
    static void Main(string[] args)
    {
        // Initialize linked list with values 1, 2, 2, 3
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(2);
        head.next.next.next = new ListNode(2);
        head.next.next.next.next = new ListNode(3);
        head.next.next.next.next.next = new ListNode(3);
        head.next.next.next.next.next.next
            = new ListNode(3);
        Console.WriteLine(
            "Before removing duplicates linked list is:");
        ListNode beforePrinter = head;
        while (beforePrinter != null) {
            Console.Write(beforePrinter.val + " ");
            beforePrinter = beforePrinter.next;
        }
        Console.WriteLine();

        Solution solution = new Solution();
        // Remove duplicates from the linked list using the
        // DeleteDuplicates function
        head = solution.DeleteDuplicates(head);

        // Print the linked list after removing duplicates
        Console.WriteLine(
            "after removing duplicates linked list is:");
        ListNode curr = head;
        while (curr != null) {
            Console.Write(curr.val + " ");
            curr = curr.next;
        }
        Console.WriteLine();
    }
}
// this code is contributed by snehalsalokhe

Javascript

// javascript code addition

// Define a node in a linked list
class Node {

      constructor(x) {
        this.val = x; // Value of the node.
        this.next = null; // Pointer to the next in the linked list.
      }
}

class Solution {

    // Remove duplicates from sorted linked list using set
    deleteDuplicates(head) {
        // Return head if it's empty
        if (head == null) return head;

        // Use set to store unique values in linked list
        let set = new Set();

        let curr = head; // Pointer to traverse the linked list
        let prev = null; // Pointer to keep track of the previous node

        // Iterate through the linked list
        while (curr) {
            // If the current value already exists in set, remove the node
            if (set.has(curr.val)) {
                prev.next = curr.next;
            } else {
                // Otherwise, add the value to set and move on to the next node
                set.add(curr.val);
                prev = curr;
            }
            curr = curr.next;
        }

        // Return the head of the linked list
        return head;
    }
};


// Initialize linked list with values 1, 2, 2, 3
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(2);
head.next.next.next = new Node(2);
head.next.next.next.next = new Node(3);
head.next.next.next.next.next = new Node(3);
head.next.next.next.next.next.next = new Node(3);

let BeforePrinter = head;

console.log("Before removing duplicates linked list is:");
console.log("\n");
while (BeforePrinter) {
    console.log(BeforePrinter.val + " ");
    BeforePrinter = BeforePrinter.next;
}
console.log("\n");

let solution = new Solution();

// Remove duplicates from the linked list using the deleteDuplicates function
head = solution.deleteDuplicates(head);

// Print the linked list after removing duplicates
console.log("after removing duplicates linked list is:");
let curr = head;
console.log("\n");
while (curr) {
    console.log(curr.val + " ");
    curr = curr.next;
}
console.log("\n");
console.log();

// This code is contributed by Nidhi goel.

Java

import java.util.*;

class ListNode {
    int val; // Value of the node
    ListNode
        next; // Pointer to the next node in the linked list

    ListNode(int x)
    {
        this.val = x;
        this.next = null;
    }
}

class Solution {
    // Remove duplicates from sorted linked list using set
    public ListNode deleteDuplicates(ListNode head)
    {
        // Return head if it's empty
        if (head == null)
            return head;

        // Use set to store unique values in linked list
        Set<Integer> set = new HashSet<>();
        ListNode curr
            = head; // Pointer to traverse the linked list
        ListNode prev = null; // Pointer to keep track of
                              // the previous node

        // Iterate through the linked list
        while (curr != null) {
            // If the current value already exists in set,
            // remove the node
            if (set.contains(curr.val)) {
                prev.next = curr.next;
            }
            else {
                // Otherwise, add the value to set and move
                // on to the next node
                set.add(curr.val);
                prev = curr;
            }
            curr = curr.next;
        }

        // Return the head of the linked list
        return head;
    }
}

public class Main {
    public static void main(String[] args)
    {
        // Initialize linked list with values 1, 2, 2, 3
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(2);
        head.next.next.next = new ListNode(2);
        head.next.next.next.next = new ListNode(3);
        head.next.next.next.next.next = new ListNode(3);
        head.next.next.next.next.next.next
            = new ListNode(3);
        ListNode BeforePrinter = head;
        System.out.println(
            "Before removing duplicates linked list is:");
        while (BeforePrinter != null) {
            System.out.print(BeforePrinter.val + " ");
            BeforePrinter = BeforePrinter.next;
        }
        System.out.println();

        Solution solution = new Solution();
        // Remove duplicates from the linked list using the
        // deleteDuplicates function
        head = solution.deleteDuplicates(head);

        // Print the linked list after removing duplicates
        System.out.println(
            "After removing duplicates linked list is:");
        ListNode curr = head;
        while (curr != null) {
            System.out.print(curr.val + " ");
            curr = curr.next;
        }
        System.out.println();
    }
}

Output

Before removing duplicates linked list is:
1 2 2 2 3 3 3 
after removing duplicates linked list is:
1 2 3 

Time Complexity: O(n)
Auxiliary Space: O(n)

Java

import java.util.HashSet;

// Definition for singly-linked list
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) { val = x; }
}

class Solution {
    // Function to remove duplicates from a sorted linked
    // list
    public ListNode deleteDuplicates(ListNode head)
    {
        // Return head if it's null
        if (head == null)
            return head;

        // Use a HashSet to store unique values in the
        // linked list
        HashSet<Integer> set = new HashSet<>();
        ListNode prev
            = head; // Keep track of the previous node
        ListNode curr = head.next; // Pointer to traverse
                                   // the linked list

        // Add the first node to the set
        set.add(head.val);

        // Iterate through the linked list
        while (curr != null) {
            // If the current value already exists in set,
            // remove the node
            if (set.contains(curr.val)) {
                prev.next = curr.next;
            }
            else {
                // Otherwise, add the value to set and move
                // on to the next node
                set.add(curr.val);
                prev = curr;
            }
            curr = curr.next;
        }

        // Return the head of the linked list
        return head;
    }
}

public class Main {
    public static void main(String[] args)
    {
        // Initialize linked list with values 1, 2, 2, 3
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(2);
        head.next.next.next = new ListNode(2);
        head.next.next.next.next = new ListNode(3);
        head.next.next.next.next.next = new ListNode(3);
        head.next.next.next.next.next.next
            = new ListNode(3);

        Solution solution = new Solution();
        ListNode Beforeprinter = head;
        System.out.print(
            "before removing the duplicates linked list is:");
        while (Beforeprinter != null) {
            System.out.print(Beforeprinter.val + " ");
            Beforeprinter = Beforeprinter.next;
        }
        System.out.println();
        // Remove duplicates from the linked list using the
        // deleteDuplicates function
        head = solution.deleteDuplicates(head);

        // Print the linked list after removing duplicates
        System.out.print(
            "after removing the duplicates linked list is:");
        ListNode curr = head;
        while (curr != null) {
            System.out.print(curr.val + " ");
            curr = curr.next;
        }
        System.out.println();
    }
}
// This Code is contributed by Veerendra Singh Rajpoot

Python3

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def delete_duplicates(self, head: ListNode) -> ListNode:

        # Return head if it's null
        if not head:
            return head

        # Use a set to store unique values in the linked list
        s = set()
        prev = head  # Keep track of the previous node
        curr = head.next  # Pointer to traverse the linked list

        # Add the first node to the set
        s.add(head.val)

        # Iterate through the linked list
        while curr:
            # If the current value already exists in set, remove the node
            if curr.val in s:
                prev.next = curr.next
            else:
                # Otherwise, add the value to set and move on to the next node
                s.add(curr.val)
                prev = curr

            curr = curr.next

        # Return the head of the linked list
        return head


if __name__ == '__main__':
    # Initialize linked list with values 1, 2, 2, 3
    head = ListNode(1)
    head.next = ListNode(2)
    head.next.next = ListNode(2)
    head.next.next.next = ListNode(2)
    head.next.next.next.next = ListNode(3)
    head.next.next.next.next.next = ListNode(3)
    head.next.next.next.next.next.next = ListNode(3)

    solution = Solution()

    Beforeprinter = head
    print("before removing the duplicates linked list is:", end=" ")
    while Beforeprinter:
        print(Beforeprinter.val, end=" ")
        Beforeprinter = Beforeprinter.next
    print()

    # Remove duplicates from the linked list using the delete_duplicates function
    head = solution.delete_duplicates(head)

    # Print the linked list after removing duplicates
    print("after removing the duplicates linked list is:", end=" ")
    curr = head
    while curr:
        print(curr.val, end=" ")
        curr = curr.next
    print()

Javascript

// Javascript code addition

// Definition for singly-linked list
class ListNode {

    constructor(val){
        this.val = val;
        this.next = null;
    }
}

class Solution {
    // Function to remove duplicates from a sorted linked list
    deleteDuplicates(head) {
        // Return head if it's null
        if (head == null) return head;

        // Use a HashSet to store unique values in the linked list
        let set = new Set();
        let prev = head; // Keep track of the previous node
        let curr = head.next; // Pointer to traverse the linked list

        // Add the first node to the set
        set.add(head.val);

        // Iterate through the linked list
        while (curr != null) {
            // If the current value already exists in set, remove the node
            if (set.has(curr.val) == true) {
                prev.next = curr.next;
            } else {
                // Otherwise, add the value to set and move on to the next node
                set.add(curr.val);
                prev = curr;
            }
            curr = curr.next;
        }

    // Return the head of the linked list
    return head;
  }
}


// Initialize linked list with values 1, 2, 2, 3
let head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(2);
head.next.next.next = new ListNode(2);
head.next.next.next.next = new ListNode(3);
head.next.next.next.next.next = new ListNode(3);
head.next.next.next.next.next.next = new ListNode(3);

let solution = new Solution();
let Beforeprinter = head;
process.stdout.write("before removing the duplicates linked list is: ");
while (Beforeprinter != null) {
    process.stdout.write(Beforeprinter.val + " ");
    Beforeprinter = Beforeprinter.next;
}
process.stdout.write("\n");
// Remove duplicates from the linked list using the deleteDuplicates function
head = solution.deleteDuplicates(head);

// Print the linked list after removing duplicates
process.stdout.write("after removing the duplicates linked list is: ");
let curr = head;
while (curr != null) {
    process.stdout.write(curr.val + " ");
    curr = curr.next;
}

//This Code is contributed by Nidhi goel.

C++

#include <iostream>
#include <unordered_set>
using namespace std;

// Definition for singly-linked list
struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x)
        : val(x)
        , next(NULL)
    {
    }
};

class Solution {
public:
    // Function to remove duplicates from a sorted linked
    // list
    ListNode* deleteDuplicates(ListNode* head)
    {
        // Return head if it's null
        if (head == NULL)
            return head;

        // Use an unordered_set to store unique values in
        // the linked list
        unordered_set<int> set;
        ListNode* prev
            = head; // Keep track of the previous node
        ListNode* curr = head->next; // Pointer to traverse
                                     // the linked list

        // Add the first node to the set
        set.insert(head->val);

        // Iterate through the linked list
        while (curr != NULL) {
            // If the current value already exists in set,
            // remove the node
            if (set.find(curr->val) != set.end()) {
                prev->next = curr->next;
            }
            else {
                // Otherwise, add the value to set and move
                // on to the next node
                set.insert(curr->val);
                prev = curr;
            }
            curr = curr->next;
        }

        // Return the head of the linked list
        return head;
    }
};

int main()
{
    // Initialize linked list with values 1, 2, 2, 3
    ListNode* head = new ListNode(1);
    head->next = new ListNode(2);
    head->next->next = new ListNode(2);
    head->next->next->next = new ListNode(2);
    head->next->next->next->next = new ListNode(3);
    head->next->next->next->next->next = new ListNode(3);
    head->next->next->next->next->next->next
        = new ListNode(3);

    Solution solution;
    ListNode* Beforeprinter = head;
    cout << "before removing the duplicates linked list "
            "is: ";
    while (Beforeprinter != NULL) {
        cout << Beforeprinter->val << " ";
        Beforeprinter = Beforeprinter->next;
    }
    cout << endl;

    // Remove duplicates from the linked list using the
    // deleteDuplicates function
    head = solution.deleteDuplicates(head);

    // Print the linked list after removing duplicates
    cout
        << "after removing the duplicates linked list is: ";
    ListNode* curr = head;
    while (curr != NULL) {
        cout << curr->val << " ";
        curr = curr->next;
    }
    cout << endl;

    return 0;
}

C#

using System;
using System.Collections.Generic;

public class ListNode {
    public int val;
    public ListNode next;
    public ListNode(int val = 0, ListNode next = null)
    {
        this.val = val;
        this.next = next;
    }
}

public class Solution {
    public ListNode DeleteDuplicates(ListNode head)
    {
        if (head == null) {
            return head;
        }
        HashSet<int> set = new HashSet<int>();
        ListNode prev = head;
        ListNode curr = head.next;

        set.Add(head.val);

        while (curr != null) {
            if (set.Contains(curr.val)) {
                prev.next = curr.next;
            }
            else {
                set.Add(curr.val);
                prev = curr;
            }

            curr = curr.next;
        }

        return head;
    }
}

public class Program {
    static void Main(string[] args)
    {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(2);
        head.next.next.next = new ListNode(2);
        head.next.next.next.next = new ListNode(3);
        head.next.next.next.next.next = new ListNode(3);
        head.next.next.next.next.next.next
            = new ListNode(3);
        Solution solution = new Solution();

        ListNode beforePrinter = head;
        Console.Write(
            "Before removing the duplicates, linked list is: ");
        while (beforePrinter != null) {
            Console.Write(beforePrinter.val + " ");
            beforePrinter = beforePrinter.next;
        }
        Console.WriteLine();

        head = solution.DeleteDuplicates(head);

        Console.Write(
            "After removing the duplicates, linked list is: ");
        ListNode curr = head;
        while (curr != null) {
            Console.Write(curr.val + " ");
            curr = curr.next;
        }
        Console.WriteLine();
    }
}
// This code is contributed by sarojmcy2e

Output

before removing the duplicates linked list is:1 2 2 2 3 3 3 
after removing the duplicates linked list is:1 2 3 

Time complexity: O(n), where n is the number of nodes in the linked list. This is because we iterate through the linked list once, and each operation of checking the set and inserting into the set takes O(logn) time on average due to the underlying balanced binary search tree data structure.

Auxiliary Space: O(n), where n is the number of unique values in the linked list. This is because we use a set to store the unique values, which takes O(n) space.

This approach is contributed by Veerendra Singh Rajpoot

If you find anything wrong or incorrect in this approach please let me know.