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

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 listusing 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 listusing 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 approachusing 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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