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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.

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

Circular linked list after deleting node from middle of the list

Program to delete a new node from the middle of the circular linked 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

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

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

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#

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

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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