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Javascript Program to Rotate Linked List block wise

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  • Last Updated : 11 Jan, 2022

Given a Linked List of length n and block length k rotate in a circular manner towards right/left each block by a number d. If d is positive rotate towards right else rotate towards left.

Examples: 

Input: 1->2->3->4->5->6->7->8->9->NULL, 
        k = 3 
        d = 1
Output: 3->1->2->6->4->5->9->7->8->NULL
Explanation: Here blocks of size 3 are
rotated towards right(as d is positive) 
by 1.
 
Input: 1->2->3->4->5->6->7->8->9->10->
               11->12->13->14->15->NULL, 
        k = 4 
        d = -1
Output: 2->3->4->1->6->7->8->5->10->11
             ->12->9->14->15->13->NULL
Explanation: Here, at the end of linked 
list, remaining nodes are less than k, i.e.
only three nodes are left while k is 4. 
Rotate those 3 nodes also by d.

Prerequisite: Rotate a linked list
The idea is if the absolute value of d is greater than the value of k, then rotate the link list by d % k times. If d is 0, no need to rotate the linked list at all. 

Javascript




<script>
// JavaScript program to rotate a
// linked list block wise 
  
// Link list node 
class Node 
{
    constructor() 
   {
        this.data = 0;
        this.next = null;
    }
}
  
var tail;
  
// Recursive function to rotate one block
function rotateHelper(blockHead,  
                      blockTail, d, k)
{
    if (d == 0)
        return blockHead;
  
    // Rotate Clockwise
    if (d > 0) 
    {
        var temp = blockHead;
        for (i = 1; temp.next.next != null &&
             i < k - 1; i++)
            temp = temp.next;
        blockTail.next = blockHead;
        tail = temp;
        return rotateHelper(blockTail, 
        temp, d - 1, k);
    }
  
    // Rotate anti-Clockwise
    if (d < 0) 
    {
        blockTail.next = blockHead;
        tail = blockHead;
        return rotateHelper(blockHead.next, 
        blockHead, d + 1, k);
    }
    return blockHead;
}
  
// Function to rotate the linked list 
// block-wise
function rotateByBlocks(head, k, d) 
{
    // If length is 0 or 1 return head
    if (head == null || head.next == null)
        return head;
  
    // if degree of rotation is 0, 
    // return head
    if (d == 0)
        return head;
  
    var temp = head;
    tail = null;
  
    // Traverse upto last element of 
    // this block
    var i;
    for (i = 1; temp.next != null && 
         i < k; i++)
        temp = temp.next;
  
    // Storing the first node of next block
    var nextBlock = temp.next;
  
    // If nodes of this block are less than k.
    // Rotate this block also
    if (i < k)
        head = rotateHelper(head, temp,
                            d % k, i);
    else
        head = rotateHelper(head, temp, 
                            d % k, k);
  
    // Append the new head of next block to
    // the tail of this block
    tail.next = rotateByBlocks(nextBlock, 
                               k, d % k);
  
    // return head of updated Linked List
    return head;
}
  
// UTILITY FUNCTIONS
// Function to push a node 
function push(head_ref, new_data) 
{
    var new_node = new Node();
    new_node.data = new_data;
    new_node.next = head_ref;
    head_ref = new_node;
    return head_ref;
}
  
// Function to print linked list 
function printList(node) 
{
    while (node != null
    {
        document.write(node.data + " ");
        node = node.next;
    }
}
  
// Driver code 
// Start with the empty list 
var head = null;
  
// Create a list 1.2.3.4.5.
// 6.7.8.9.null
for (i = 9; i > 0; i -= 1)
    head = push(head, i);
document.write("Given linked list <br/>");
printList(head);
  
// k is block size and d is number of
w// rotations in every block.
var k = 3, d = 2;
head = rotateByBlocks(head, k, d);
document.write(
"<br/>Rotated by blocks Linked list <br/>");
printList(head);
// This code is contributed by gauravrajput1 
</script>

Output:

Given linked list 
1 2 3 4 5 6 7 8 9 
Rotated by blocks Linked list 
2 3 1 5 6 4 8 9 7

Please refer complete article on Rotate Linked List block wise for more details!


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