Java Program to Delete a Node from the Beginning of the Circular Linked List

In this DSA problem, we will learn to create a circular linked list and delete the node from the beginning of that.

The circular linked list connects the last node with the first node. To remove the first node from the linked list, we can make the second node a root node, and connect the last node with the second node.

Problem statement

We have given a circular linked list. We need to delete the starting node of the linked list.

Sample examples

Input

1 -> 2 -> 3 -> 4 -> 8 -> 10 

Output

2 -> 3 -> 4 -> 8 -> 10

Explanation

We have deleted the node from starting.

Input

100 -> 103 -> 200 -> 99 -> 56 -> 78 -> 532

Output

103 -> 200 -> 99 -> 56 -> 78 -> 532

Explanation

The first node from the circular linked list has been deleted.

Approach 1

In this approach, we will create the circular linked list first. While adding the node to the linked list, we will connect the last node with the root node. Furthermore, while deleting the first node, we will make the next node of the root node a root node and connect the last node with the updated root node.

Algorithm

• Step 1 − Create a 'listNode' class containing the 'value' of the integer data type and the' next' listNode pointer. Also, define the constructor to initialize the 'value' variable.

• Step 2 − Next, define the 'root' and 'last' pointers to store the first and last node of the linked list.

• Step 3 − Now, define the addValue() function to add the node to the linked list.

• Step 3.1 − Create a new listNode with the value we got as a parameter.

• Step 3.2 − If the root node is NULL, assign a new node to the 'root' and 'last' node. Also, connect the new node with the root node.

• Step 3.3 − If the root node is not NULL, connect the new node with the last node, update the value of the 'last' pointer, and connect the last node with the root node.

• Step 4 − Define the deleteFirstNode() function to delete the node from the start.

• Step 4.1 − In the deleteFirstNode() function, If the root is null, execute the return statement.

• Step 4.2 − If the root and last node are same, update the root and last with NULL.

• Step 4.3 − If the root and last node are not the same, update the root node with its next node, and connect the last node with the updated root node.

• Step 5 − Define the showList() function to print the list values.

• Step 5.1 − Initialize the 'temp' node with the root node.

• Step 5.2 − If the root is NULL, print the message that 'List is empty'.

• Step 5.3 − Otherwise, traverse the list, and print the value of the 'temp' node. Also, update the 'temp' node with its next node.

• Step 6 − In the main() method, create the object of the Main class.

• Step 7 − By taking the object of Main class as a reference, execute the addValue() method to add nodes to the linked list. After that, call the showList() to show the original list.

• Step 8 − Next, execute the deleteFirstNode() function to delete the first node, and call the showList() method to show the updated linked list.

Example

public class Main {

   // Node for creating the linked list
   public class listNode {
      int value;
      listNode next;
      
      // Constructor
      public listNode(int val) {
         this.value = val;
      }
   }
   
   // Root and last node initialization
   public listNode root = null;
   public listNode last = null;
   
   // Method to add new Node
   public void addValue(int val) {
   
      // Cerate new listNode
      listNode newNode = new listNode(val);
      
      // For the first node
      if (root == null) {
         root = newNode;
         last = newNode;
         newNode.next = root;
      }
      
      // Add new node after the last node
      // New node points to the root node
      else {
         last.next = newNode;
         last = newNode;
         last.next = root;
      }
   }
   public void deleteFirstNode() {
   
      // For an empty list
      if (root == null) {
         return;
      }
      
      // Root node points to the next node. The last node points to the updated root node
      else {
         if (root != last) {
            root = root.next;
            last.next = root;
         }
         
         // For a list having a single node
         else {
            root = last = null;
         }
      }
   }
   
   // displaying the nodes
   public void showList() {
      listNode temp = root;
      if (root == null) {
         System.out.println("The list is empty");
      } else {
      
         // Traverse the list to show each node's value
         do {
            System.out.print(" " + temp.value);
            temp = temp.next;
         } while (temp != root);
            System.out.println();
      }
   }
   public static void main(String[] args) {
      Main list = new Main();
      
      // Adds data to the list
      list.addValue(1);
      list.addValue(2);
      list.addValue(3);
      list.addValue(4);
      list.addValue(8);
      list.addValue(10);
      
      // Printing the original list
      System.out.println("The initial list is: - ");
      list.showList();
      
      // Delete the first node
      list.deleteFirstNode();
      System.out.println("After deleting the first node, the list is: - ");
      list.showList();
      
      // Delete the second node
      list.deleteFirstNode();
      System.out.println("After deleting the second node, the list is: - ");
      list.showList();
    }
}

Output

The initial list is: - 
 1 2 3 4 8 10
After deleting the first node, the list is: - 
 2 3 4 8 10
After deleting the second node, the list is: - 
 3 4 8 10

• Time complexity − O(1), as we modify the pointer in the constant time.

• Space complexity − O(1) as we don't use dynamic space.

We learned to delete the first node from the circular linked list. The programmer may learn to delete the last node of the circular linked list. In this case, programmers need to connect the last-second element of the linked list with the root node.