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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.

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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:

Program to rotate doubly linked list by N nodes

List after rotating it by 3 nodes:

Program to rotate doubly linked list by N 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.
  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. 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:
    #Represent the head and tail of the doubly linked list
    def __init__(self):
        self.head = None;
        self.tail = None;
        self.size = 0;
        
    #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;
        #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
        current = self.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
            self.tail.next = self.head;
            #Node next to nth node will be new head
            self.head = current.next;
            #Previous node to head should be None
            self.head.previous = 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
        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 = RotateList();
#Add nodes to the list
dList.addNode(1);
dList.addNode(2);
dList.addNode(3);
dList.addNode(4);
dList.addNode(5);
 
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;
//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;
    }
    //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
    struct node *current = 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
        tail->next = head;
        //Node next to nth node will be new head
        head = current->next;
        //Previous node to head should be NULL
        head->previous = 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
    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(4);
    addNode(5);
    
    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;
    //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;
        }
        //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
        Node current = 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
            tail.next = head;
            //Node next to nth node will be new head
            head = current.next;
            //Previous node to head should be null
            head.previous = 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
        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) {
        
        RotateList dList = new RotateList();
        //Add nodes to the list
        dList.addNode(1);
        dList.addNode(2);
        dList.addNode(3);
        dList.addNode(4);
        dList.addNode(5);
        
        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; 
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 RotateList<T>{
            int size = 0;
            //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;
                }
                //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
                Node<T> current = 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
                    tail.next = head;
                    //Node next to nth node will be new head
                    head = current.next;
                    //Previous node to head should be null
                    head.previous = 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
                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()
        {
            RotateList<int> dList = new RotateList<int>();
            //Add nodes to the list
            dList.addNode(1);
            dList.addNode(2);
            dList.addNode(3);
            dList.addNode(4);
            dList.addNode(5);
 
            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{
    //Represent the head and tail of the doubly linked list
    public $head;
    public $tail;
    public $size = 0;
    function __construct(){
        $this->head = NULL;
        $this->tail = NULL;
        $this->size = 0;
    }
    
    //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;
        }
        //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
        $current = $this->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
            $this->tail->next = $this->head;
            //Node next to nth node will be new head
            $this->head = $current->next;
            //Previous node to head should be NULL
            $this->head->previous = 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
        $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 RotateList();
//Add nodes to the list
$dList->addNode(1);
$dList->addNode(2);
$dList->addNode(3);
$dList->addNode(4);
$dList->addNode(5);
 
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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