# Program to Remove Duplicate Elements From a Doubly Linked List

Program to Remove Duplicate Elements From a Doubly Linked List on fibonacci, factorial, prime, armstrong, swap, reverse, search, sort, stack, queue, array, linkedlist, tree, graph etc.

## Q. Program to remove duplicate elements from a doubly linked list.

### 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.
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.

## Solution

### 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:
def __init__(self):
self.tail = None;

#Create a new node
newNode = Node(data);

#If list is empty
#Both head and tail will point to newNode
#head's previous will point to 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
return;
else:
#Initially, current will point to head node
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
print("List is empty");
return;
while(current != None):
#Prints each node by incrementing pointer.
print(current.data),
current = current.next;
print();

dList = RemoveDuplicate();

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;
};

struct node *head, *tail = NULL;

//Create a new node
struct node *newNode = (struct node*)malloc(sizeof(struct node));
newNode->data = data;

//If list is empty
//Both head and tail will point to newNode
//head's previous will point to 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
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
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()
{

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;
}
}

//Create a new node
Node newNode = new Node(data);

//If list is empty
//Both head and tail will point to newNode
//head's previous will point to 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
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
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();

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;
{
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>{
protected Node<T> tail = null;

//Create a new node
Node<T> newNode = new Node<T>(data);

//If list is empty
//Both head and tail will point to newNode
//head's previous will point to 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
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
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>();

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{
public \$tail;
function __construct(){
\$this->tail = NULL;
}

//Create a new node
\$newNode = new Node(\$data);

//If list is empty
//Both head and tail will point to newNode
//head's previous will point to 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
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
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();

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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