# Program to Delete a New Node From the Middle of the Circular Linked List

Program to Delete a New Node From the Middle of the Circular Linked List on fibonacci, factorial, prime, armstrong, swap, reverse, search, sort, stack, queue, array, linkedlist, tree, graph etc.

## Q. Program to delete a new node from the middle of the circular linked list.

### Explanation

In 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 mid-point 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 mid-point 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. Mid-point of the node is 2. To remove C from the list, we will iterate through the list till mid-point. Now current will point to B and temp will point to C. C will be removed when B will point to D.

### Algorithm

1. Define a Node class which represents a node in the list. It has two properties data and next which will point to the next node.
2. Define another class for creating the circular linked list and it has two nodes: head and tail. It has a variable size and two methods: deleteMid() and display() .
3. deleteMid() will delete the node from the middle of the list:
1. It first checks whethe the head is null (empty list) then, it will return from the function as there is no node present in the list.
2. If the list is not empty, it will check whether the list has only one node.
3. If the list has only one node, it will set both head and tail to null.
4. If the list has more than one node then, it will calculate the size of the list. Divide the size by 2 and store it in the variable count.
5. Temp will point to head, and current will point to the previous node to temp.
6. Iterate the list till current will point middle node of the list.
7. Current will point to node next to temp, i.e., removes the node next to current.
4. 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 will points to head again.
3. Current will point to the next node in the list in each iteration.

## Solution

### Python

```#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.tail = Node(None);
self.size = 0;

#This function will add the new node at the end of the list.
newNode = Node(data);
#Checks if the list is empty.
#If list is empty, both head and tail would point to new node.
self.tail = newNode;
else:
#tail will point to new node.
self.tail.next = newNode;
#New node will become new tail.
self.tail = newNode;
#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
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.
#Initially temp will point to 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, count-1):
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:
#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.size = self.size - 1;

#Displays all the nodes in the list
def display(self):
print("List is empty");
return;
else:
#Prints each node by incrementing pointer.
print(current.data),
current = current.next;
print(current.data),
print("\n");

cl = CreateList();
#Printing original list
print("Original List:");
cl.display();
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 *tail = NULL;

//This function will add the new node at the end of the list.
//Create new node
struct node *newNode = (struct node*)malloc(sizeof(struct node));
newNode->data = data;
//Checks if the list is empty.
//If list is empty, both head and tail would point to new node.
tail = newNode;
}else {
//tail will point to new node.
tail->next = newNode;
//New node will become new tail.
tail = newNode;
}
//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
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;
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 < count-1; 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 {
//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 {
}
}
size--;
}

//This function will display the nodes of circular linked list
void display(){
printf("List is empty");
}
else{
do{
//Prints each node by incrementing pointer.
printf("%d ", current->data);
current = current->next;
printf("\n");
}
}

int main()
{
//Printing original list
printf("Original List:\n ");
display();
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
```

### JAVA

```public 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 tail = null;

//This function will add the new node at the end of the list.
//Create new node
Node newNode = new Node(data);
//Checks if the list is empty.
//If list is empty, both head and tail would point to new node.
tail = newNode;
}
else {
//tail will point to new node.
tail.next = newNode;
//New node will become new tail.
tail = newNode;
}
//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
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;
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 < count-1; 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 {
//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 {
}
}
size--;
}

//Displays all the nodes in the list
public void display() {
System.out.println("List is empty");
}
else {
do{
//Prints each node by incrementing pointer.
System.out.print(" "+ current.data);
current = current.next;
System.out.println();
}
}

public static void main(String[] args) {
DeleteMid cl = new DeleteMid();
//Printing original list
System.out.println("Original List: ");
cl.display();
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;
{
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> tail = null;

//This function will add the new node at the end of the list.
//Create new node
Node<T> newNode = new Node<T>(data);
//Checks if the list is empty.
tail = newNode;
}else{
//tail will point to new node.
tail.next = newNode;
//New node will become new tail.
tail = newNode;
}
//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
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;
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 < count-1; 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 {
//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 {
}
}
size--;
}

//Displays all the nodes in the list
public void display(){
Console.WriteLine("List is empty");
}
else{
do{
//Prints each node by incrementing pointer.
Console.Write(" "+ current.data);
current = current.next;
Console.WriteLine();
}
}
}

public static void Main()
{

CreateList<int> cl = new CreateList<int>();
//Printing original list
Console.WriteLine("Original List: ");
cl.display();
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 \$tail;
function __construct(){
\$this->tail = NULL;
\$this->size = 0;
}
//This function will add the new node at the end of the list.
//Create new node
\$newNode = new Node(\$data);
//Checks if the list is empty.
//If list is empty, both head and tail would point to new node.
\$this->tail = \$newNode;
}
else{
//tail will point to new node.
\$this->tail->next = \$newNode;
//New node will become new tail.
\$this->tail = \$newNode;
}
\$this->size++;
}

//Deletes node from the middle of the list
function deleteMid() {
//Checks whether list is empty
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;
\$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 < \$count-1; \$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 {
//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->size--;
}

//Displays all the nodes in the list
function display() {
echo "List is empty";
}
else {
do{
//Prints each node by incrementing pointer.
echo(" \$current->data");
\$current = \$current->next;
echo "<br>";
}
}
}
\$cl = new CreateList();
//Printing original list
echo "Original List:<br>";
\$cl->display();
\$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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