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Q. Program to delete a new node from the middle of the circular linked list.ExplanationIn this program, we will create a circular linked list and delete a node from the middle of the list. If the list is empty, display the message "List is empty". If the list is not empty, we will calculate the size of the list and then divide it by 2 to get the midpoint of the list. We maintain two pointers temp and current. Current will point to the previous node of temp. We will iterate through the list until midpoint is reached then the current will point to the middle node. We delete middle node such that current's next will be temp's next node. Circular linked list after deleting node from middle of the list Consider the above list. Size of the list is 4. Midpoint of the node is 2. To remove C from the list, we will iterate through the list till midpoint. Now current will point to B and temp will point to C. C will be removed when B will point to D. 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; self.size = 0; #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; #Size will count the number of nodes in the list self.size = int(self.size)+1; #Deletes node from the middle of the list def deleteMid(self): #Checks whether list is empty if(self.head == None): return; else: #Store the mid position of the list count = (self.size//2) if (self.size % 2 == 0) else ((self.size+1)//2); #Checks whether head is equal to tail or not, if yes then list has only one node. if( self.head != self.tail ): #Initially temp will point to head; temp = self.head; current = None; #Current will point to node previous to temp #If temp is pointing to node 2 then current will points to node 1. for i in range(0, count1): current = temp; temp = temp.next; if(current != None): #temp is the middle that needs to be removed. #So, current node will point to node next to temp by skipping temp. current.next = temp.next; #Delete temp; temp = None; #Current points to null then head and tail will point to node next to temp. else: self.head = self.tail = temp.next; self.tail.next = self.head; #Delete temp; temp = None; #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; self.size = self.size  1; #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.deleteMid(); #Printing updated list print("Updated List:"); cl.display(); Output: Original List: 1 2 3 4 Updated List: 1 3 4 Updated List: 1 4 Updated List: 4 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; }; int size; //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; } //Counts the number of nodes in list size++; } //Deletes node from the middle of the list void deleteMid() { struct node *current, *temp; //Checks whether list is empty if(head == NULL) { return; } else { //Store the mid position of the list int count = (size % 2 == 0) ? (size/2) : ((size+1)/2); //Checks whether head is equal to tail or not, if yes then list has only one node. if( head != tail ) { //Initially temp will point to head; temp = head; current = NULL; //Current will point to node previous to temp //If temp is pointing to node 2 then current will points to node 1. for(int i = 0; i < count1; i++){ current = temp; temp = temp>next; } if(current != NULL) { //temp is the middle that needs to be removed. //So, current node will point to node next to temp by skipping temp. current>next = temp>next; //Delete temp; temp = NULL; } //Current points to null then head and tail will point to node next to temp. else { head = tail = temp>next; tail>next = head; //Delete temp; temp = NULL; } } //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; } } size; } //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) { deleteMid(); //Printing updated list printf("Updated List:\n "); display(); } return 0; } Output: Original List: 1 2 3 4 Updated List: 1 3 4 Updated List: 1 4 Updated List: 4 Updated List: List is empty JAVApublic class DeleteMid { //Represents the node of list. public class Node{ int data; Node next; public Node(int data) { this.data = data; } } public int size; //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; } //Counts the number of nodes in list size++; } //Deletes node from the middle of the list public void deleteMid() { Node current, temp; //Checks whether list is empty if(head == null) { return; } else { //Store the mid position of the list int count = (size % 2 == 0) ? (size/2) : ((size+1)/2); //Checks whether head is equal to tail or not, if yes then list has only one node. if( head != tail ) { //Initially temp will point to head; temp = head; current = null; //Current will point to node previous to temp //If temp is pointing to node 2 then current will points to node 1. for(int i = 0; i < count1; i++){ current = temp; temp = temp.next; } if(current != null) { //temp is the middle that needs to be removed. //So, current node will point to node next to temp by skipping temp. current.next = temp.next; //Delete temp; temp = null; } //Current points to null then head and tail will point to node next to temp. else { head = tail = temp.next; tail.next = head; //Delete temp; temp = null; } } //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; } } size; } //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) { DeleteMid cl = new DeleteMid(); //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.deleteMid(); //Printing updated list System.out.println("Updated List: "); cl.display(); } } } Output: Original List: 1 2 3 4 Updated List: 1 3 4 Updated List: 1 4 Updated List: 4 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 int size = 0; 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; } //Counts the number of nodes in the list size++; } //Deletes node from the middle of the list public void deleteMid() { Node<T> current, temp; //Checks whether list is empty if(head == null) { return; } else { //Store the mid position of the list int count = (size % 2 == 0) ? (size/2) : ((size+1)/2); //Checks whether head is equal to tail or not, if yes then list has only one node. if( head != tail ) { //Initially temp will point to head; temp = head; current = null; //Current will point to node previous to temp //If temp is pointing to node 2 then current will points to node 1. for(int i = 0; i < count1; i++){ current = temp; temp = temp.next; } if(current != null) { //temp is the middle that needs to be removed. //So, current node will point to node next to temp by skipping temp. current.next = temp.next; //Delete temp; temp = null; } //Current points to null then head and tail will point to node next to temp. else { head = tail = temp.next; tail.next = head; //Delete temp; temp = null; } } //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; } } size; } //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.deleteMid(); //Printing updated list Console.WriteLine("Updated List: "); cl.display(); } } } } Output: Original List: 1 2 3 4 Updated List: 1 3 4 Updated List: 1 4 Updated List: 4 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; $this>size = 0; } //This function will add the new node at the end of the list. function add($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; } $this>size++; } //Deletes node from the middle of the list function deleteMid() { //Checks whether list is empty if($this>head == NULL) { return; } else { //Store the mid position of the list $count = ($this>size % 2 == 0) ? ($this>size/2) : (($this>size+1)/2); //Checks whether head is equal to tail or not, if yes then list has only one node. if( $this>head != $this>tail ) { //Initially temp will point to head; $temp = $this>head; $current = NULL; //Current will point to node previous to temp //If temp is pointing to node 2 then current will points to node 1. for($i = 0; $i < $count1; $i++){ $current = $temp; $temp = $temp>next; } if($current != NULL) { //temp is the middle that needs to be removed. //So, current node will point to node next to temp by skipping temp. $current>next = $temp>next; //Delete temp; $temp = NULL; } //Current points to null then head and tail will point to node next to temp. else { $this>head = $this>tail = $temp>next; $this>tail>next = $this>head; //Delete temp; $temp = NULL; } } //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; } } $this>size; } //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>deleteMid(); //Printing updated list echo "Updated List:<br>"; $cl>display(); } ?> </body> </html> Output: Original List: 1 2 3 4 Updated List: 1 3 4 Updated List: 1 4 Updated List: 4 Updated List: List is empty
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