# C# Program For Removing Duplicates From A Sorted Linked List

• Last Updated : 14 Dec, 2021

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``using` `System;``     ` `public` `class` `LinkedList``{``    ``// Head of list``    ``Node head; `` ` `    ``// 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;``            ``}``  ` `            ``// Advance if no deletion``            ``else` `                ``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 `

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.

Recursive Approach :

## 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 off the new node ``        ``new_node.next = (head_ref);     ``         ` `        ``// Move the head to point to the ``        ``// new node ``        ``(head_ref) = new_node; ``        ``return` `head_ref;``    ``} `` ` `    ``/* Function to print nodes in ``       ``a given linked list */``    ``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(``"Linked list after"` `+ ``                      ``" duplicate removal "``);     ``        ``printList(head);             ``    ``}``}``// This code is contributed by PrinciRaj1992`

Output:

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

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#

 `// 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`

Output:

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

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#

 `// 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 off ``        ``// the new node``        ``new_node.next = (head_ref);``          ` `        ``/* Move the head to point``           ``to the new node */``        ``head_ref = new_node;``        ``return` `head_ref;``    ``}``      ` `    ``/* Function to print nodes ``       ``in a given linked list */``    ``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(``                ``"Linked list after duplicate removal  "``);``        ``removeDuplicates(head);``    ``}``}``// This code is contributed by rag2127`

Output:

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

Time Complexity: O(Number of Nodes)

Space Complexity: O(Number of Nodes)

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

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