Reverse a Linked List in groups of given size ‘K’

Objective: Given a linked list and integer ‘k’, write an algorithm to reverse the linked list in groups of size ‘k’.

Example:

Reverse a Linked List in groups of given size 'K' Example
Reverse a Linked List in groups of given size ‘K’ Example

Approach:

  • Earlier we have seen how to reverse a linked list, solution for reverse the linked list in groups of size will be extension of this solution.
  • Reverse first ‘k’ nodes of the linked list, the kth node will be a new head, return it.
  • Make a recursive call to rest of the list and attach it to the last node.(See the picture below)
Reverse a Linked List in groups of given size 'K'
Reverse a Linked List in groups of given size ‘K’

Complete Code:


public class ReverseInGrps {
public Node reveseGrps(Node head, int k){
int x = k;
Node head_next=null;
Node h = head;
Node head_prev = null;
while(h!=null && x>0){
head_next = h.next;
h.next = head_prev;
head_prev = h;
h = head_next;
x;
}
if(head_next!=null){
head.next = reveseGrps(head_next,k);
}
return head_prev;
}
public void display(Node head){
Node n=head;
while(n!=null){
System.out.print("->" + n.data);
n=n.next;
}
}
public static void main(String[] args) {
Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
head.next.next.next.next.next = new Node(6);
head.next.next.next.next.next.next = new Node(7);
head.next.next.next.next.next.next.next = new Node(8);
ReverseInGrps r = new ReverseInGrps();
Node x = r.reveseGrps(head, 3);
r.display(x);
}
}
class Node{
int data;
Node next;
public Node(int data){
this.data = data;
}
}

Output:

->3->2->1->6->5->4->8->7

1 thought on “Reverse a Linked List in groups of given size ‘K’”

  1. Hi, this code wont work, if say i have a linked list like 1->2->3->4->5 ans i want to reverse in group of 3 nodes then it will return 3->2->1->5->4 which is not correct. Can you suggest me another approach where we get 3->2->1->4->5 only.Thanks.

    Reply

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