Given a linked list of n nodes which is first sorted, then rotated by k elements. Find the value of k.
The idea is to traverse singly linked list to check condition whether current node value is greater than value of next node. If the given condition is true, then break the loop. Otherwise increase the counter variable and increase the node by node->next. Below is the implementation of this approach.
Java
// Program for count number of rotations in// sorted linked list.import java.util.*;class GFG{ /* Linked list node */static class Node { int data; Node next;};// Function that count number of// rotation in singly linked list.static int countRotation(Node head){ // declare count variable and assign it 1. int count = 0; // declare a min variable and assign to // data of head node. int min = head.data; // check that while head not equal to null. while (head != null) { // if min value is greater then head.data // then it breaks the while loop and // return the value of count. if (min > head.data) break; count++; // head assign the next value of head. head = head.next; } return count;}// Function to push element in linked list.static Node push(Node head, int data){ // Allocate dynamic memory for newNode. Node newNode = new Node(); // Assign the data into newNode. newNode.data = data; // newNode.next assign the address of // head node. newNode.next = (head); // newNode become the headNode. (head) = newNode; return head;}// Display linked list.static void printList(Node node){ while (node != null) { System.out.printf("%d ", node.data); node = node.next; }}// Driver functionspublic static void main(String[] args){ // Create a node and initialize with null Node head = null; // push() insert node in linked list. // 15.18.5.8.11.12 head = push(head, 12); head = push(head, 11); head = push(head, 8); head = push(head, 5); head = push(head, 18); head = push(head, 15); printList(head); System.out.println(); System.out.print("Linked list rotated elements: "); // Function call countRotation() System.out.print(countRotation(head) +"");}}// This code contributed by gauravrajput1 |
Output
15 18 5 8 11 12 Linked list rotated elements: 2
Time Complexity: O(N), where N represents the length of the linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Count rotations in sorted and rotated linked list for more details!
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