# Program to Rotate Doubly Linked List by N Nodes

Program to Rotate Doubly Linked List by N Nodes on fibonacci, factorial, prime, armstrong, swap, reverse, search, sort, stack, queue, array, linkedlist, tree, graph etc.

## Q. Program to rotate doubly linked list by N nodes.

### Explanation

In this program, we need to create a doubly linked list and rotate it by n node. This can be achieved by maintaining a pointer that starts from the head node and traverses the list until current points to the nth node. Move the list from head to the nth node and place it after tail. Now nth node will be the tail of the list and node next to nth node will be the new head. Here, n should always be greater than 0 but less than the size of the list.

Original List:

List after rotating it by 3 nodes:

In the above example, we need to rotate list by 3 nodes. First, we iterate through the list until current points to the 3rd node which is, in this case, are node 3. Move the list from node 1 to 3 and place it after tail. Now, node 4 will be new head and node 3 will be the new tail.

### 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 the 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. rotateList() will rotate the list by given n nodes.
1. First, check whether n is 0 or greater than or equal to many nodes present in the list.
2. If yes, print the list as it is.
3. If no, define a node current which will point to head.
4. Iterate through the list till current reaches the nth node.
5. Tail's next will point to head node.
6. Make node next to current as the new head. Head's previous will point to null.
7. The current node will become tail of the list. Tail's next will point 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 RotateList:
def __init__(self):
self.tail = None;
self.size = 0;

#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;
#Size will count the number of nodes present in the list
self.size = self.size + 1;

#rotateList() will rotate the list by given n nodes
def rotateList(self, n):
#Initially, current will point to head

#n should not be 0 or greater than or equal to number of nodes present in the list
if(n == 0 or n >= self.size):
return;
else:
#Traverse through the list till current point to nth node
#after this loop, current will point to nth node
for i in range(1, n):
current = current.next;

#Now to move entire list from head to nth node and add it after tail
#Node next to nth node will be new head
#Previous node to head should be None
#nth node will become new tail of the list
self.tail = current;
#tail's next will point to None
self.tail.next = None;

#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 = RotateList();

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

#Rotates list by 3 nodes
dList.rotateList(3);

print("Updated List: ");
dList.display();
```

Output:

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

### C

```#include <stdio.h>

//Represent a node of the doubly linked list

struct node{
int data;
struct node *previous;
struct node *next;
};

int size = 0;
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;
}
//Size will count the number of nodes present in the list
size++;
}

//rotateList() will rotate the list by given n nodes
void rotateList(int n) {
//Initially, current will point to head

//n should not be 0 or greater than or equal to number of nodes present in the list
if(n == 0 || n >= size)
return;
else {
//Traverse through the list till current point to nth node
//after this loop, current will point to nth node
for(int i = 1; i < n; i++)
current = current->next;

//Now to move entire list from head to nth node and add it after tail
//Node next to nth node will be new head
//Previous node to head should be NULL
//nth node will become new tail of the list
tail = current;
//tail's next will point to NULL
tail->next = 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("Original List: \n");
display();

//Rotates list by 3 nodes
rotateList(3);

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

return 0;
}
```

Output:

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

### JAVA

```public class RotateList {

//Represent a node of the doubly linked list

class Node{
int data;
Node previous;
Node next;

public Node(int data) {
this.data = data;
}
}

int size = 0;

//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;
}
//Size will count the number of nodes present in the list
size++;
}

//rotateList() will rotate the list by given n nodes
public void rotateList(int n) {
//Initially, current will point to head

//n should not be 0 or greater than or equal to number of nodes present in the list
if(n == 0 || n >= size)
return;
else {
//Traverse through the list till current point to nth node
//after this loop, current will point to nth node
for(int i = 1; i < n; i++)
current = current.next;

//Now to move entire list from head to nth node and add it after tail
//Node next to nth node will be new head
//Previous node to head should be null
//nth node will become new tail of the list
tail = current;
//tail's next will point to null
tail.next = 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) {

RotateList dList = new RotateList();

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

//Rotates list by 3 nodes
dList.rotateList(3);

System.out.println("Updated List: ");
dList.display();
}
}
```

Output:

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

### 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 RotateList<T>{
int size = 0;
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;
}
//Size will count the number of nodes present in the list
size++;
}

//rotateList() will rotate the list by given n nodes
public void rotateList(int n) {
//Initially, current will point to head

//n should not be 0 or greater than or equal to number of nodes present in the list
if(n == 0 || n >= size)
return;
else {
//Traverse through the list till current point to nth node
//after this loop, current will point to nth node
for(int i = 1; i < n; i++)
current = current.next;

//Now to move entire list from head to nth node and add it after tail
//Node next to nth node will be new head
//Previous node to head should be null
//nth node will become new tail of the list
tail = current;
//tail's next will point to null
tail.next = 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()
{
RotateList<int> dList = new RotateList<int>();

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

//Rotates list by 3 nodes
dList.rotateList(3);

Console.WriteLine("Updated List: ");
dList.display();
}
}
}
```

Output:

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

### 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 RotateList{
public \$tail;
public \$size = 0;
function __construct(){
\$this->tail = NULL;
\$this->size = 0;
}

//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;
}
//Size will count the number of nodes present in the list
\$this->size++;
}

//rotateList() will rotate the list by given n nodes
function rotateList(\$n) {
//Initially, current will point to head

//n should not be 0 or greater than or equal to number of nodes present in the list
if(\$n == 0 || \$n >= \$this->size)
return;
else {
//Traverse through the list till current point to nth node
//after this loop, current will point to nth node
for(\$i = 1; \$i < \$n; \$i++)
\$current = \$current->next;

//Now to move entire list from head to nth node and add it after tail
//Node next to nth node will be new head
//Previous node to head should be NULL
//nth node will become new tail of the list
\$this->tail = \$current;
//tail's next will point to NULL
\$this->tail->next = 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 RotateList();

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

//Rotates list by 3 nodes
\$dList->rotateList(3);

print("Updated List: <br>");
\$dList->display();
?>
</body>
</html>
```

Output:

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

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