In this article, we are going to learn to delete the middle node from the circular Linked List in java. The approach we are going to follow for this program is, first we calculate the number of nodes in the list and then divide the number of nodes by 2 to get the middle node of the list.
Algorithm
Case 1: List is empty
- Then simply just return from the function.
Case 2: List has only one node
- It means head == tail then assigned head and tail equal to null.
Case 3: List has only two nodes
- It means the middle node of the list at 1 position. In this case, we have to modify the head and tail also. Head becomes tail and tail.next becomes also tail.
Case 4: List has more than two nodes
- It means the position of the middle node is greater than 1. In this case, we define the two-node temp and prev. Assign head to the temp.
- Then traverse the list using temp up to the middle node. In each iteration assign prev = temp and temp = temp.next
- Now temp is pointing to the middle node and prev pointing to the previous node of the middle node.
- Then simply set prev.next=temp.next and temp=null,
Java
// Java Program to Delete a// Node From the Middle of the// Circular Linked Listclass CLinkedList { class Node { int data; Node next; } private static int size; // head points first node and // tail points to last node private Node head, tail; CLinkedList() { this.head = null; this.tail = null; size = 0; } public void addNode(int d) { // create a empty node Node n = new Node(); // list is empty if (this.head == null) { n.data = d; this.head = n; this.tail = n; n.next = this.head; } // list has one node or more than one nodes else { n.data = d; tail.next = n; tail = n; tail.next = head; } size++; } public void deleteNodeMiddle() { int loc; Node temp, prev; // calculating position of middle node if (size % 2 == 0) { loc = size / 2; } else { loc = (size / 2) + 1; } // list is empty if (head == null) { return; } // list contains only one node else if (head == tail) { head = null; tail = null; } // list contains only two nodes else if (loc == 1) { head = tail; tail.next = tail; } // list contains more than two nodes else { temp = head; prev = null; int i = 1; while (i < loc) { prev = temp; temp = temp.next; i++; } prev.next = temp.next; temp = null; } size--; if (size < 0) { size = 0; } } public void display() { // displaying list if (head == null) { System.out.println("List is empty"); } else { Node temp = head; do { System.out.print(temp.data + " "); temp = temp.next; } while (temp != head); System.out.println(); } }}class Test { public static void main(String args[]) { CLinkedList c1 = new CLinkedList(); // adding node in linkedlist c1.addNode(10); c1.addNode(20); c1.addNode(30); c1.addNode(40); // displaying nodes after adding System.out.print("List="); c1.display(); // delete node in middle c1.deleteNodeMiddle(); // displaying linkedlist after deleting middle node System.out.print( "List after deleting middle node="); c1.display(); c1.deleteNodeMiddle(); // displaying linkedlist after deleting middle node System.out.print( "List after deleting middle node="); c1.display(); c1.deleteNodeMiddle(); // displaying linkedlist after deleting middle node System.out.print( "List after deleting middle node="); c1.display(); c1.deleteNodeMiddle(); // displaying linkedlist after deleting middle node System.out.print( "List after deleting middle node="); c1.display(); }} |
Output
List=10 20 30 40 List after deleting middle node=10 30 40 List after deleting middle node=10 40 List after deleting middle node=40 List after deleting middle node=List is empty
Time complexity: O(N) where N is no of nodes in given circular linked list
Auxiliary space: O(1)
