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Java Program For Reversing A Linked List In Groups Of Given Size – Set 1

  • Last Updated : 16 Dec, 2021

Given a linked list, write a function to reverse every k nodes (where k is an input to the function). 

Example: 

Input: 1->2->3->4->5->6->7->8->NULL, K = 3 
Output: 3->2->1->6->5->4->8->7->NULL 
Input: 1->2->3->4->5->6->7->8->NULL, K = 5 
Output: 5->4->3->2->1->8->7->6->NULL 

Algorithm: reverse(head, k) 

  • Reverse the first sub-list of size k. While reversing keep track of the next node and previous node. Let the pointer to the next node be next and pointer to the previous node be prev. See this post for reversing a linked list.
  • head->next = reverse(next, k) ( Recursively call for rest of the list and link the two sub-lists )
  • Return prev ( prev becomes the new head of the list (see the diagrams of an iterative method of this post )

Below is image shows how the reverse function works: 

Below is the implementation of the above approach:

Java




// Java program to reverse a 
// linked list in groups of
// given size
class LinkedList
{
    // Head of list
    Node head; 
  
    // Linked list Node
    class Node 
    {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
  
    Node reverse(Node head, int k)
    {
        if(head == null)
          return null;
        Node current = head;
        Node next = null;
        Node prev = null;
  
        int count = 0;
  
        // Reverse first k nodes of 
        // linked list
        while (count < k && 
               current != null
        {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
            count++;
        }
  
        /* next is now a pointer to (k+1)th node
           Recursively call for the list starting from
           current. And make rest of the list as next of
           first node */
        if (next != null)
            head.next = reverse(next, k);
  
        // prev is now head of the input list
        return prev;
    }
  
    // Utility functions
  
    // Inserts a new Node at front
    // of the list.
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);
  
        // 3. Make next of new Node 
        // as head 
        new_node.next = head;
  
        // 4. Move the head to point to
        // new Node
        head = new_node;
    }
  
    // Function to print linked list
    void printList()
    {
        Node temp = head;
        while (temp != null
        {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }
  
    // Driver code
    public static void main(String args[])
    {
        LinkedList llist = new LinkedList();
  
        // Create Linked List is 
        // 1->2->3->4->5->6->
        // 7->8->8->9->null
        llist.push(9);
        llist.push(8);
        llist.push(7);
        llist.push(6);
        llist.push(5);
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);
  
        System.out.println("Given Linked List");
        llist.printList();
  
        llist.head = llist.reverse(llist.head, 3);
  
        System.out.println("Reversed list");
        llist.printList();
    }
}
// This code is contributed by Rajat Mishra 

Output: 

Given Linked List
1 2 3 4 5 6 7 8 9 
Reversed list
3 2 1 6 5 4 9 8 7 

Complexity Analysis: 

  • Time Complexity: O(n). 
    Traversal of list is done only once and it has ‘n’ elements.
  • Auxiliary Space: O(n/k). 
    For each Linked List of size n, n/k or (n/k)+1 calls will be made during the recursion.

Please refer complete article on Reverse a Linked List in groups of given size | Set 1 for more details!


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