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Q. Program to delete a new node from the end of the circular linked list.ExplanationIn this program, we will create a circular linked list and delete a node from the end of the list. If the list is empty, it will display the message "List is empty". If the list is not empty, we will loop through the list till second last node is reached. We will make second last node as the new tail, and this new tail will point to head and delete the previous tail. Circular linked list after deleting node from end Here, in the above list, D is the last node which needs to be deleted. We will iterate through the list till C. Make C as new tail and C will point back to head A. Algorithm
SolutionPython#Represents the node of list. class Node: def __init__(self,data): self.data = data; self.next = None; class CreateList: #Declaring head and tail pointer as null. def __init__(self): self.head = Node(None); self.tail = Node(None); self.head.next = self.tail; self.tail.next = self.head; #This function will add the new node at the end of the list. def add(self,data): newNode = Node(data); #Checks if the list is empty. if self.head.data is None: #If list is empty, both head and tail would point to new node. self.head = newNode; self.tail = newNode; newNode.next = self.head; else: #tail will point to new node. self.tail.next = newNode; #New node will become new tail. self.tail = newNode; #Since, it is circular linked list tail will point to head. self.tail.next = self.head; #Deletes node from end of the list def deleteEnd(self): #Checks whether list is empty if(self.head == None): return; else: #Checks whether contain only one element if(self.head != self.tail ): current = self.head; #Loop will iterate till the second last element as current.next is pointing to tail while(current.next != self.tail): current = current.next; #Second last element will be new tail self.tail = current; #Tail will point to head as it is a circular linked list self.tail.next = self.head; #If the list contains only one element #Then it will remove it and both head and tail will point to null else: self.head = self.tail = None; #Displays all the nodes in the list def display(self): current = self.head; if self.head is None: print("List is empty"); return; else: #Prints each node by incrementing pointer. print(current.data), while(current.next != self.head): current = current.next; print(current.data), print("\n"); class CircularLinkedList: cl = CreateList(); #Adds data to the list cl.add(1); cl.add(2); cl.add(3); cl.add(4); #Printing original list print("Original List:"); cl.display(); while(cl.head != None): cl.deleteEnd(); #Printing updated list print("Updated List:"); cl.display(); Output: Original List: 1 2 3 4 Updated List: 1 2 3 Updated List: 1 2 Updated List: 1 Updated List: List is empty C#include <stdio.h> #include <string.h> #include <stdlib.h> //Represents the node of list. struct node{ int data; struct node *next; }; //Declaring head and tail pointer as null. struct node *head = NULL; struct node *tail = NULL; //This function will add the new node at the end of the list. void add(int data){ //Create new node struct node *newNode = (struct node*)malloc(sizeof(struct node)); newNode->data = data; //Checks if the list is empty. if(head == NULL){ //If list is empty, both head and tail would point to new node. head = newNode; tail = newNode; newNode->next = head; }else { //tail will point to new node. tail->next = newNode; //New node will become new tail. tail = newNode; //Since, it is circular linked list tail will point to head. tail->next = head; } } //Deletes node from end of the list void deleteEnd() { //Checks whether list is empty if(head == NULL) { return; } else { //Checks whether contain only one element if(head != tail ) { struct node *current = head; //Loop will iterate till the second last element as current.next is pointing to tail while(current->next != tail) { current = current->next; } //Second last element will be new tail tail = current; //Tail will point to head as it is a circular linked list tail->next = head; } //If the list contains only one element //Then it will remove it and both head and tail will point to null else { head = tail = NULL; } } } //This function will display the nodes of circular linked list void display(){ struct node *current = head; if(head == NULL){ printf("List is empty"); } else{ do{ //Prints each node by incrementing pointer. printf("%d ", current->data); current = current->next; }while(current != head); printf("\n"); } } int main() { //Adds data to the list add(1); add(2); add(3); add(4); //Printing original list printf("Original List:\n "); display(); while(head != NULL) { deleteEnd(); //Printing updated list printf("Updated List:\n "); display(); } return 0; } Output: Original List: 1 2 3 4 Updated List: 1 2 3 Updated List: 1 2 Updated List: 1 Updated List: List is empty JAVApublic class DeleteEnd { //Represents the node of list. public class Node{ int data; Node next; public Node(int data) { this.data = data; } } //Declaring head and tail pointer as null. public Node head = null; public Node tail = null; //This function will add the new node at the end of the list. public void add(int data){ //Create new node Node newNode = new Node(data); //Checks if the list is empty. if(head == null) { //If list is empty, both head and tail would point to new node. head = newNode; tail = newNode; newNode.next = head; } else { //tail will point to new node. tail.next = newNode; //New node will become new tail. tail = newNode; //Since, it is circular linked list tail will point to head. tail.next = head; } } //Deletes node from end of the list public void deleteEnd() { //Checks whether list is empty if(head == null) { return; } else { //Checks whether contain only one element if(head != tail ) { Node current = head; //Loop will iterate till the second last element as current.next is pointing to tail while(current.next != tail) { current = current.next; } //Second last element will be new tail tail = current; //Tail will point to head as it is a circular linked list tail.next = head; } //If the list contains only one element //Then it will remove it and both head and tail will point to null else { head = tail = null; } } } //Displays all the nodes in the list public void display() { Node current = head; if(head == null) { System.out.println("List is empty"); } else { do{ //Prints each node by incrementing pointer. System.out.print(" "+ current.data); current = current.next; }while(current != head); System.out.println(); } } public static void main(String[] args) { DeleteEnd cl = new DeleteEnd(); //Adds data to the list cl.add(1); cl.add(2); cl.add(3); cl.add(4); //Printing original list System.out.println("Original List: "); cl.display(); while(cl.head != null) { cl.deleteEnd(); //Printing updated list System.out.println("Updated List: "); cl.display(); } } } Output: Original List: 1 2 3 4 Updated List: 1 2 3 Updated List: 1 2 Updated List: 1 Updated List: List is empty C#using System; namespace CircularLinkedList { public class Program { //Represents the node of list. public class Node<T>{ public T data; public Node<T> next; public Node(T value) { data = value; next = null; } } public class CreateList<T>{ public Node<T> head = null; public Node<T> tail = null; //This function will add the new node at the end of the list. public void add(T data){ //Create new node Node<T> newNode = new Node<T>(data); //Checks if the list is empty. if(head == null){ head = newNode; tail = newNode; newNode.next = head; }else{ //tail will point to new node. tail.next = newNode; //New node will become new tail. tail = newNode; //Since, it is circular linked list tail will point to head. tail.next = head; } } //Deletes node from end of the list public void deleteEnd() { //Checks whether list is empty if(head == null) { return; } else { //Checks whether contain only one element if(head != tail ) { Node<T> current = head; //Loop will iterate till the second last element as current.next is pointing to tail while(current.next != tail) { current = current.next; } //Second last element will be new tail tail = current; //Tail will point to head as it is a circular linked list tail.next = head; } //If the list contains only one element //Then it will remove it and both head and tail will point to null else { head = tail = null; } } } //Displays all the nodes in the list public void display(){ Node<T> current = head; if(head == null){ Console.WriteLine("List is empty"); } else{ do{ //Prints each node by incrementing pointer. Console.Write(" "+ current.data); current = current.next; }while(current != head); Console.WriteLine(); } } } public static void Main() { CreateList<int> cl = new CreateList<int>(); //Adds data to the list cl.add(1); cl.add(2); cl.add(3); cl.add(4); //Printing original list Console.WriteLine("Original List: "); cl.display(); while(cl.head != null) { cl.deleteEnd(); //Printing updated list Console.WriteLine("Updated List: "); cl.display(); } } } } Output: Original List: 1 2 3 4 Updated List: 1 2 3 Updated List: 1 2 Updated List: 1 Updated List: List is empty PHP<!DOCTYPE html> <html> <body> <?php //Represents the node of list. class Node{ public $data; public $next; function __construct($data){ $this->data = $data; $this->next = NULL; } } class CreateList{ //Declaring head and tail pointer as null. public $head; public $tail; function __construct(){ $this->head = NULL; $this->tail = NULL; } //This function will add the new node at the end of the list. function add($data){ //Create new node $newNode = new Node($data); //Checks if the list is empty. if($this->head == NULL){ //If list is empty, both head and tail would point to new node. $this->head = $newNode; $this->tail = $newNode; $newNode->next = $this->head; } else{ //tail will point to new node. $this->tail->next = $newNode; //New node will become new tail. $this->tail = $newNode; //Since, it is circular linked list tail will point to head. $this->tail->next = $this->head; } } //Deletes node from end of the list function deleteEnd() { //Checks whether list is empty if($this->head == NULL) { return; } else { //Checks whether contain only one element if($this->head != $this->tail ) { $current = $this->head; //Loop will iterate till the second last element as current.next is pointing to tail while($current->next != $this->tail) { $current = $current->next; } //Second last element will be new tail $this->tail = $current; //Tail will point to head as it is a circular linked list $this->tail->next = $this->head; } //If the list contains only one element //Then it will remove it and both head and tail will point to null else { $this->head = $this->tail = NULL; } } } //Displays all the nodes in the list function display() { $current = $this->head; if($this->head == NULL) { echo "List is empty"; } else { do{ //Prints each node by incrementing pointer. echo(" $current->data"); $current = $current->next; }while($current != $this->head); echo "<br>"; } } } $cl = new CreateList(); //Adds data to the list $cl->add(1); $cl->add(2); $cl->add(3); $cl->add(4); //Printing original list echo "Original List:<br>"; $cl->display(); while($cl->head != NULL) { $cl->deleteEnd(); //Printing updated list echo "Updated List:<br>"; $cl->display(); } ?> </body> </html> Output: Original List: 1 2 3 4 Updated List: 1 2 3 Updated List: 1 2 Updated List: 1 Updated List: List is empty
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