# Program to Create a Doubly Linked List of N Nodes and Display it in Reverse Order

Program to Create a Doubly Linked List of N Nodes and Display it in Reverse Order on fibonacci, factorial, prime, armstrong, swap, reverse, search, sort, stack, queue, array, linkedlist, tree, graph etc.

## Q. Program to create a doubly linked list of n nodes and display it in reverse order.

### Explanation

In this program, we create a doubly linked list, then reverse the list by reversing the direction of the list and print out the nodes.

Traverse through the list by swapping the previous pointer with next pointer of each node. Then, swap the position of head and tail node that is, head of the original list will become tail of new list and tail of the original list will become head of the new list. So, the reversed list will be:

### 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. reverse() will reverse the given doubly linked list.
1. Define a node current which will initially point to head.
2. Traverse through the list by making current to point to current.next in each iteration till current points to null.
3. In each iteration, swap previous and next pointer of each node to reverse the direction of the list.
4. In the end, swap the position of head and tail.
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 ReverseList:
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;

#reverse() will reverse the doubly linked list
def reverse(self):
#Node current will point to head

#Swap the previous and next pointers of each node to reverse the direction of the list
while(current != None):
temp = current.next;
current.next = current.previous;
current.previous = temp;
current = current.previous;
#Swap the head and tail pointers.
self.tail = temp;

#display() will print out the elements 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;

dList = ReverseList();

print("Original List: ");
dList.display();

#Reverse the given list
dList.reverse();

#Displays the reversed list
print("\nReversed List: ");
dList.display();
```

Output:

```Original List:
1 2 3 4 5
Reversed List:
5 4 3 2 1
```

### 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;
}
}

//reverse() will reverse the doubly linked list
void reverse() {
//Node current will point to head
struct node *current = head, *temp = NULL;

//Swap the previous and next pointers of each node to reverse the direction of the list
while(current != NULL) {
temp = current->next;
current->next = current->previous;
current->previous = temp;
current = current->previous;
}
//Swap the head and tail pointers.
tail = temp;
}

//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;
}
}

int main()
{

printf("Original List: \n");
display();

//Reverse the given list
reverse();

//Displays the reversed list
printf("\nReversed List: \n");
display();

return 0;
}
```

Output:

```Original List:
1 2 3 4 5
Reversed List:
5 4 3 2 1
```

### JAVA

```public class ReverseList {

//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;
}
}

//reverse() will reverse the doubly linked list
public void reverse() {
//Node current will point to head
Node current = head, temp = null;

//Swap the previous and next pointers of each node to reverse the direction of the list
while(current != null) {
temp = current.next;
current.next = current.previous;
current.previous = temp;
current = current.previous;
}
//Swap the head and tail pointers.
tail = temp;
}

//display() will print out the elements 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;
}
}

public static void main(String[] args) {

ReverseList dList = new ReverseList();

System.out.println("Original List: ");
dList.display();

//Reverse the given list
dList.reverse();

//Displays the reversed list
System.out.println("\nReversed List: ");
dList.display();
}
}
```

Output:

```Original List:
1 2 3 4 5
Reversed List:
5 4 3 2 1
```

### 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 ReverseList<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;
}
}

//reverse() will reverse the doubly linked list
public void reverse() {
//Node current will point to head
Node<T> current = head, temp = null;

//Swap the previous and next pointers of each node to reverse the direction of the list
while(current != null) {
temp = current.next;
current.next = current.previous;
current.previous = temp;
current = current.previous;
}
//Swap the head and tail pointers.
tail = temp;
}

//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;
}
}
}

public static void Main()
{
ReverseList<int> dList = new ReverseList<int>();

Console.WriteLine("Original List: ");
dList.display();

//Reverse the given list
dList.reverse();

//Displays the reversed list
Console.WriteLine("\nReversed List: ");
dList.display();
}
}
}
```

Output:

```Original List:
1 2 3 4 5
Reversed List:
5 4 3 2 1
```

### 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 ReverseList{
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;
}
}

//reverse() will reverse the doubly linked list
function reverse() {
//Node current will point to head

//Swap the previous and next pointers of each node to reverse the direction of the list
while(\$current != NULL) {
\$temp = \$current->next;
\$current->next = \$current->previous;
\$current->previous = \$temp;
\$current = \$current->previous;
}
//Swap the head and tail pointers.
\$this->tail = \$temp;
}

//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;
}
}
}

\$dList = new ReverseList();

print("Original List: <br>");
\$dList->display();

//Reverse the given list
\$dList->reverse();

//Displays the reversed list
print("<br>Reversed List: <br>");
\$dList->display();
?>
</body>
</html>
```

Output:

```Original List:
1 2 3 4 5
Reversed List:
5 4 3 2 1
```

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