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# 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 ` `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 ` `#include `   `/* 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

 ``

Output

```Linked list before duplicate removal
11 11 11 13 13 20
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 ` `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 ` `#include `   `/* 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

 ``

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 ` `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 ` `#include `   `/* 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

 ``

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 ` `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 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

 ``

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.

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 ` `#include `   `// 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 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 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 ` `#include ` `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.

My Personal Notes arrow_drop_up