Java Program to Count rotations in sorted and rotated linked list
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
import java.util.*;
class GFG
{
static class Node {
int data;
Node next;
};
static int countRotation(Node head)
{
int count = 0 ;
int min = head.data;
while (head != null ) {
if (min > head.data)
break ;
count++;
head = head.next;
}
return count;
}
static Node push(Node head, int data)
{
Node newNode = new Node();
newNode.data = data;
newNode.next = (head);
(head) = newNode;
return head;
}
static void printList(Node node)
{
while (node != null ) {
System.out.printf( "%d " , node.data);
node = node.next;
}
}
public static void main(String[] args)
{
Node head = null ;
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: " );
System.out.print(countRotation(head) +"
");
}
}
|
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!
Last Updated :
26 May, 2022
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