In a circular linked list, every node points to its next node in the sequence but the last node points to the first node in the list. Here, Create a circular linked list and sort the circular linked list in ascending order.
Circular linked list before sorting:
CIRCULAR LINKED LIST
Circular linked list after sorting:
SORTED CIRCULAR LINKED LIST
Approach:
- Take two pointers: Current pointing to head of the node and Temp pointing to next node of Current.
- Now for each iteration compare value of Current pointer to the value of Temp pointer.
Here two cases arise
Case 1: If the value of a current pointer is greater than that of Temp pointer
Swap values of a current pointer and temp pointer.
Move the temp pointer to next node
Case 2: If the value of a current pointer is less than or equal to that of Temp pointer
Move the temp pointer to next node
- Now keep doing this until temp.next !=head of the list.
- After completing step 3 move the Current to next node and repeat the steps 1,2,3 .
- Each iteration results in fixing of the shortest element of the list to it’s correct position.
- Repeat the above steps until Current. Next != head of list .
Let’s see how this work for the first node of the given circular linked list
Below is the implementation of the above approach:
Java
// Java Program to Sort the Elements// of the Circular Linked Listimport java.io.*;public class GFG { // Stores Information about Node of List public class Node { int data; Node next; public Node(int data) { this.data = data; } } // Declaring Head of the Node public Node head_of_node = null; // A last pointer to help append values to our list public Node last = null; // Add method adds values to the end of the list public void add(int data) { Node newNode = new Node(data); if (head_of_node == null) { head_of_node = newNode; last = newNode; newNode.next = head_of_node; } else { last.next = newNode; last = newNode; last.next = head_of_node; } } // Sort_List method sorts the circular // linked list Using the algorithm public void Sort_List() { // current pointer pointing to the head of the list Node current = head_of_node; // a temp pointer Node temp = null; // variable value helps in swap of the values int value; // this is the Algorithm discussed above if (head_of_node == null) { System.out.println("Your list is empty"); } else { while (current.next != head_of_node) { temp = current.next; while (temp != head_of_node) { if (current.data > temp.data) { value = current.data; current.data = temp.data; temp.data = value; } temp = temp.next; } current = current.next; } } } // Print_list method iterates through the list and // prints the values stored in the list public void Print_List() { Node current = head_of_node; if (head_of_node == null) { System.out.println("Your list is empty"); } else { do { System.out.print(" " + current.data); current = current.next; } while (current != head_of_node); System.out.println(); } } // Driver code public static void main(String[] args) { GFG circular_list = new GFG(); circular_list.add(10); circular_list.add(6); circular_list.add(3); circular_list.add(8); circular_list.add(4); System.out.print("Original List --> "); circular_list.Print_List(); circular_list.Sort_List(); System.out.print("List after Sorting--> "); circular_list.Print_List(); }} |
Output
Original List --> 10 6 3 8 4 List after Sorting--> 3 4 6 8 10
Time Complexity: O(N2)
Auxiliary Space: O(1) as it is using constant space
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