Q:

Program to remove duplicate elements from a doubly linked list

belongs to collection: Doubly Linked List Programs

0

Explanation

In this program, we will create a doubly linked list and remove the duplicate, if present, by traversing through the list.

Original List:

List after removing duplicates:

In above list, node2 is repeated thrice, and node 3 is repeated twice. Current will point to head, and index will point to node next to current. Start traversing the list till a duplicate is found that is when current's data is equal to index's data. In above example, the first duplicate will be found at position 4. Assign index to another node temp. Connect index's previous node with index's next node. Delete temp which was pointing to duplicate node. This process will continue till all duplicates are removed.

Algorithm

  1. Define a Node class which represents a node in the list. It will have three properties: data, previous which will point to the previous node and next which will point to the next node.
  2. Define another class for creating a doubly linked list, and it has two nodes: head and tail. Initially, head and tail will point to null.
  3. addNode() will add node to the list:
    1. It first checks whether the head is null, then it will insert the node as the head.
    2. Both head and tail will point to a newly added node.
    3. Head's previous pointer will point to null and tail's next pointer will point to null.
    4. If the head is not null, the new node will be inserted at the end of the list such that new node's previous pointer will point to tail.
    5. The new node will become the new tail. Tail's next pointer will point to null.
  4. removeDuplicateNode() will remove duplicate nodes from the list.
    1. Define a new node current which will initially point to head.
    2. Node index will always point to node next to current.
    3. Loop through the list until current points to null.
    4. Check whether current?s data is equal to index's data that means index is duplicate of current.
    5. Node temp will point to index to store duplicate node.
    6. The previous node to the index will point to next node to index.
    7. Since, temp is pointing to index, which is a duplicate node, so set temp to null.
  5. display() will show all the nodes present in the list.
    1. Define a new node 'current' that will point to the head.
    2. Print current.data till current points to null.
    3. Current will point to the next node in the list in each iteration.

Input:

 

#Add nodes to the list  

dList.addNode(1);  

dList.addNode(2);  

dList.addNode(3);  

dList.addNode(2);  

dList.addNode(2);  

dList.addNode(4);  

dList.addNode(5);  

dList.addNode(3);  

Output:

Originals list: 1 2 3 2 2 4 5 3 
List after removing duplicates: 1 2 3 4 5 

All Answers

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Python

#Represent a node of doubly linked list  
class Node:  
    def __init__(self,data):  
        self.data = data;  
        self.previous = None;  
        self.next = None;  
          
class RemoveDuplicate:  
    #Represent the head and tail of the doubly linked list  
    def __init__(self):  
        self.head = None;  
        self.tail = None;  
          
    #addNode() will add a node to the list  
    def addNode(self, data):  
        #Create a new node  
        newNode = Node(data);  
          
        #If list is empty  
        if(self.head == None):  
            #Both head and tail will point to newNode  
            self.head = self.tail = newNode;  
            #head's previous will point to None  
            self.head.previous = None;  
            #tail's next will point to None, as it is the last node of the list  
            self.tail.next = None;  
        else:  
            #newNode will be added after tail such that tail's next will point to newNode  
            self.tail.next = newNode;  
            #newNode's previous will point to tail  
            newNode.previous = self.tail;  
            #newNode will become new tail  
            self.tail = newNode;  
            #As it is last node, tail's next will point to None  
            self.tail.next = None;  
              
    #removeDuplicateNode() will remove duplicate nodes from the list  
    def removeDuplicateNode(self):  
          
        #Checks whether list is empty  
        if(self.head == None):  
            return;  
        else:  
            #Initially, current will point to head node  
            current = self.head;  
            while(current != None):  
                #index will point to node next to current  
                index = current.next  
                while(index != None):  
                    if(current.data == index.data):  
                        #Store the duplicate node in temp  
                        temp = index;  
                        #index's previous node will point to node next to index thus, removes the duplicate node  
                        index.previous.next = index.next;  
                        if(index.next != None):  
                            index.next.previous = index.previous;  
                        #Delete duplicate node by making temp to None  
                        temp = None;  
                    index = index.next;  
                current = current.next;  
                          
    #display() will print out the nodes of the list  
    def display(self):  
        #Node current will point to head  
        current = self.head;  
        if(self.head == None):  
            print("List is empty");  
            return;  
        while(current != None):  
            #Prints each node by incrementing pointer.  
            print(current.data),  
            current = current.next;  
        print();  
          
dList = RemoveDuplicate();  
#Add nodes to the list  
dList.addNode(1);  
dList.addNode(2);  
dList.addNode(3);  
dList.addNode(2);  
dList.addNode(2);  
dList.addNode(4);  
dList.addNode(5);  
dList.addNode(3);  
   
print("Originals list: ");  
dList.display();  
   
#Removes duplicate nodes  
dList.removeDuplicateNode();  
   
print("List after removing duplicates: ");  
dList.display();  

 

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

 

C

#include <stdio.h>  
   
//Represent a node of the doubly linked list  
  
struct node{  
    int data;  
    struct node *previous;  
    struct node *next;  
};      
   
//Represent the head and tail of the doubly linked list  
struct node *head, *tail = NULL;  
   
//addNode() will add a node to the list  
void addNode(int data) {  
    //Create a new node  
    struct node *newNode = (struct node*)malloc(sizeof(struct node));  
    newNode->data = data;  
      
    //If list is empty  
    if(head == NULL) {  
        //Both head and tail will point to newNode  
        head = tail = newNode;  
        //head's previous will point to NULL  
        head->previous = NULL;  
        //tail's next will point to NULL, as it is the last node of the list  
        tail->next = NULL;  
    }  
    else {  
        //newNode will be added after tail such that tail's next will point to newNode  
        tail->next = newNode;  
        //newNode's previous will point to tail  
        newNode->previous = tail;  
        //newNode will become new tail  
        tail = newNode;  
        //As it is last node, tail's next will point to NULL  
        tail->next = NULL;  
    }  
}  
   
//removeDuplicateNode() will remove duplicate nodes from the list  
void removeDuplicateNode() {  
    //Node current will point to head  
    struct node *current, *index, *temp;  
      
    //Checks whether list is empty  
    if(head == NULL) {  
        return;  
    }  
    else {  
        //Initially, current will point to head node  
        for(current = head; current != NULL; current = current->next) {  
            //index will point to node next to current  
            for(index = current->next; index != NULL; index = index->next) {  
                if(current->data == index->data) {  
                    //Store the duplicate node in temp  
                    temp = index;  
                    //index's previous node will point to node next to index thus, removes the duplicate node  
                    index->previous->next = index->next;  
                    if(index->next != NULL)  
                        index->next->previous = index->previous;  
                    //Delete duplicate node by making temp to NULL  
                    temp = NULL;  
                }  
            }  
        }  
    }  
}  
   
//display() will print out the nodes of the list  
void display() {  
    //Node current will point to head  
    struct node *current = head;  
    if(head == NULL) {  
        printf("List is empty\n");  
        return;  
    }  
    while(current != NULL) {  
        //Prints each node by incrementing pointer.  
        printf("%d ", current->data);  
        current = current->next;  
    }  
    printf("\n");  
}  
   
int main()  
{  
    //Add nodes to the list  
    addNode(1);  
    addNode(2);  
    addNode(3);  
    addNode(2);  
    addNode(2);  
    addNode(4);  
    addNode(5);  
    addNode(3);  
      
    printf("Originals list: \n");  
    display();  
      
    //Removes duplicate nodes  
    removeDuplicateNode();  
      
    printf("List after removing duplicates: \n");  
    display();  
   
    return 0;  
}  

 

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

 

JAVA

public class RemoveDuplicate {  
      
    //Represent a node of the doubly linked list  
  
    class Node{  
        int data;  
        Node previous;  
        Node next;  
          
        public Node(int data) {  
            this.data = data;  
        }  
    }  
      
    //Represent the head and tail of the doubly linked list  
    Node head, tail = null;  
      
    //addNode() will add a node to the list  
    public void addNode(int data) {  
        //Create a new node  
        Node newNode = new Node(data);  
          
        //If list is empty  
        if(head == null) {  
            //Both head and tail will point to newNode  
            head = tail = newNode;  
            //head's previous will point to null  
            head.previous = null;  
            //tail's next will point to null, as it is the last node of the list  
            tail.next = null;  
        }  
        else {  
            //newNode will be added after tail such that tail's next will point to newNode  
            tail.next = newNode;  
            //newNode's previous will point to tail  
            newNode.previous = tail;  
            //newNode will become new tail  
            tail = newNode;  
            //As it is last node, tail's next will point to null  
            tail.next = null;  
        }  
    }  
      
    //removeDuplicateNode() will remove duplicate nodes from the list  
    public void removeDuplicateNode() {  
        //Node current will point to head  
        Node current, index, temp;  
          
        //Checks whether list is empty  
        if(head == null) {  
            return;  
        }  
        else {  
            //Initially, current will point to head node  
            for(current = head; current != null; current = current.next) {  
                //index will point to node next to current  
                for(index = current.next; index != null; index = index.next) {  
                    if(current.data == index.data) {  
                        //Store the duplicate node in temp  
                        temp = index;  
                        //index's previous node will point to node next to index thus, removes the duplicate node  
                        index.previous.next = index.next;  
                        if(index.next != null)  
                            index.next.previous = index.previous;  
                        //Delete duplicate node by making temp to null  
                        temp = null;  
                    }  
                }  
            }  
        }  
    }  
      
    //display() will print out the nodes of the list  
    public void display() {  
        //Node current will point to head  
        Node current = head;  
        if(head == null) {  
            System.out.println("List is empty");  
            return;  
        }  
        while(current != null) {  
            //Prints each node by incrementing the pointer.  
  
            System.out.print(current.data + " ");  
            current = current.next;  
        }  
        System.out.println();  
    }  
      
    public static void main(String[] args) {  
          
        RemoveDuplicate dList = new RemoveDuplicate();  
        //Add nodes to the list  
        dList.addNode(1);  
        dList.addNode(2);  
        dList.addNode(3);  
        dList.addNode(2);  
        dList.addNode(2);  
        dList.addNode(4);  
        dList.addNode(5);  
        dList.addNode(3);  
          
        System.out.println("Originals list: ");  
        dList.display();  
          
        //Removes duplicate nodes  
        dList.removeDuplicateNode();  
          
        System.out.println("List after removing duplicates: ");  
        dList.display();  
    }  
}  

 

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

 

C#

using System;   
namespace DoublyLinkedList   
{                       
    public class Program  
    {  
        //Represent a node of the doubly linked list  
  
        public class Node<T>{  
            public T data;  
            public Node<T> previous;  
            public Node<T> next;  
              
            public Node(T value) {  
                data = value;  
            }  
        }  
          
        public class RemoveDuplicate<T>{  
            //Represent the head and tail of the doubly linked list  
            protected Node<T> head = null;               
             protected Node<T> tail = null;  
              
            //addNode() will add a node to the list  
            public void addNode(T data) {  
                //Create a new node  
                Node<T> newNode = new Node<T>(data);  
   
                //If list is empty  
                if(head == null) {  
                    //Both head and tail will point to newNode  
                    head = tail = newNode;  
                    //head's previous will point to null  
                    head.previous = null;  
                    //tail's next will point to null, as it is the last node of the list  
                    tail.next = null;  
                }  
                else {  
                    //newNode will be added after tail such that tail's next will point to newNode  
                    tail.next = newNode;  
                    //newNode's previous will point to tail  
                    newNode.previous = tail;  
                    //newNode will become new tail  
                    tail = newNode;  
                    //As it is last node, tail's next will point to null  
                    tail.next = null;  
                }  
            }  
              
            //removeDuplicateNode() will remove duplicate nodes from the list  
            public void removeDuplicateNode() {  
                //Node current will point to head  
                Node<T> current, index, temp;  
   
                //Checks whether list is empty  
                if(head == null) {  
                    return;  
                }  
                else {  
                    //Initially, current will point to head node  
                    for(current = head; current != null; current = current.next) {  
                        //index will point to node next to current  
                        for(index = current.next; index != null; index = index.next) {  
                            if(current.data.Equals(index.data)) {  
                                //Store the duplicate node in temp  
                                temp = index;  
                                //index's previous node will point to node next to index thus, removes the duplicate node  
                                index.previous.next = index.next;  
                                if(index.next != null)  
                                    index.next.previous = index.previous;  
                                //Delete duplicate node by making temp to null  
                                temp = null;  
                            }  
                        }  
                    }  
                }  
            }  
      
            //display() will print out the nodes of the list  
            public void display() {  
                //Node current will point to head  
                Node<T> current = head;  
                if(head == null) {  
                    Console.WriteLine("List is empty");  
                    return;  
                }  
                while(current != null) {  
                    //Prints each node by incrementing the pointer.  
  
                    Console.Write(current.data + " ");  
                    current = current.next;  
                }  
                Console.WriteLine();  
            }  
        }  
          
        public static void Main()  
        {  
            RemoveDuplicate<int> dList = new RemoveDuplicate<int>();  
            //Add nodes to the list  
            dList.addNode(1);  
            dList.addNode(2);  
            dList.addNode(3);  
            dList.addNode(2);  
            dList.addNode(2);  
            dList.addNode(4);  
            dList.addNode(5);  
            dList.addNode(3);  
   
            Console.WriteLine("Originals list: ");  
            dList.display();  
   
            //Removes duplicate nodes  
            dList.removeDuplicateNode();  
   
            Console.WriteLine("List after removing duplicates: ");  
            dList.display();  
        }      
    }  
}  

 

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5 

 

PHP

<!DOCTYPE html>  
<html>  
<body>  
<?php  
//Represent a node of doubly linked list  
class Node{  
    public $data;  
    public $previous;  
    public $next;  
      
    function __construct($data){  
        $this->data = $data;  
    }  
}  
class RemoveDuplicate{  
    //Represent the head and tail of the doubly linked list  
    public $head;  
    public $tail;  
    function __construct(){  
        $this->head = NULL;  
        $this->tail = NULL;  
    }  
      
    //addNode() will add a node to the list  
    function addNode($data){  
        //Create a new node  
        $newNode = new Node($data);  
          
        //If list is empty  
        if($this->head == NULL) {  
            //Both head and tail will point to newNode  
            $this->head = $this->tail = $newNode;  
            //head's previous will point to NULL  
            $this->head->previous = NULL;  
            //tail's next will point to NULL, as it is the last node of the list  
            $this->tail->next = NULL;  
        }  
        else {  
            //newNode will be added after tail such that tail's next will point to newNode  
            $this->tail->next = $newNode;  
            //newNode's previous will point to tail  
            $newNode->previous = $this->tail;  
            //newNode will become new tail  
            $this->tail = $newNode;  
            //As it is last node, tail's next will point to NULL  
            $this->tail->next = NULL;  
        }  
    }  
      
    //removeDuplicateNode() will remove duplicate nodes from the list  
    function removeDuplicateNode() {          
        //Checks whether list is empty  
        if($this->head == NULL) {  
            return;  
        }  
        else {  
            //Initially, current will point to head node  
            for($current = $this->head; $current != NULL; $current = $current->next) {  
                //index will point to node next to current  
                for($index = $current->next; $index != NULL; $index = $index->next) {  
                    if($current->data == $index->data) {  
                        //Store the duplicate node in temp  
                        $temp = $index;  
                        //index's previous node will point to node next to index thus, removes the duplicate node  
                        $index->previous->next = $index->next;  
                        if($index->next != NULL)  
                            $index->next->previous = $index->previous;  
                        //Delete duplicate node by making temp to NULL  
                        $temp = NULL;  
                    }  
                }  
            }  
        }  
    }  
      
    //display() will print out the nodes of the list  
    function display() {  
        //Node current will point to head  
        $current = $this->head;  
        if($this->head == NULL) {  
            print("List is empty <br>");  
            return;  
        }  
        while($current != NULL) {  
            //Prints each node by incrementing pointer.  
            print($current->data . " ");  
            $current = $current->next;  
        }  
        print("<br>");  
    }  
}  
      
$dList = new RemoveDuplicate();  
//Add nodes to the list  
$dList->addNode(1);  
$dList->addNode(2);  
$dList->addNode(3);  
$dList->addNode(2);  
$dList->addNode(2);  
$dList->addNode(4);  
$dList->addNode(5);  
$dList->addNode(3);  
   
print("Originals list: <br>");  
$dList->display();  
   
//Removes duplicate nodes  
$dList->removeDuplicateNode();  
   
print("List after removing duplicates: <br>");  
$dList->display();  
?>  
</body>  
</html>  

 

Output:

Originals list: 
1 2 3 2 2 4 5 3 
List after removing duplicates: 
1 2 3 4 5

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total answers (1)

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