Program to Rotate Doubly Linked List by N Nodes

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

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.
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.
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.
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.
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 emptyn”);          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

Next Topic#

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

Explanation

Algorithm

Solution

Python

C

JAVA

C#

PHP

Wind Energy

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