Java Program For Reversing Alternate K Nodes In A Singly Linked List

Given a linked list, write a function to reverse every alternate k nodes (where k is an input to the function) in an efficient way. Give the complexity of your algorithm.

Example:

Inputs:   1->2->3->4->5->6->7->8->9->NULL and k = 3
Output:   3->2->1->4->5->6->9->8->7->NULL. 

Method 1 (Process 2k nodes and recursively call for rest of the list): 
This method is basically an extension of the method discussed in this post. 

kAltReverse(struct node *head, int k)
  1)  Reverse first k nodes.
  2)  In the modified list head points to the kth node.  So change next 
       of head to (k+1)th node
  3)  Move the current pointer to skip next k nodes.
  4)  Call the kAltReverse() recursively for rest of the n - 2k nodes.
  5)  Return new head of the list.

Java

// Java program to reverse alternate k
// nodes in a linked list
class LinkedList
{
    static Node head;

    class Node
    {
        int data;
        Node next;

        Node(int d)
        {
            data = d;
            next = null;
        }
    }

    /* Reverses alternate k nodes and
       returns the pointer to the new
       head node */
    Node kAltReverse(Node node, int k)
    {
        Node current = node;
        Node next = null, prev = null;
        int count = 0;

        / *1) reverse first k nodes of the
          linked list */
        while (current != null && count < k)
        {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
            count++;
        }

        /* 2) Now head points to the kth node.
           So change next of head to (k+1)th node*/
        if (node != null)
        {
            node.next = current;
        }

        /* 3) We do not want to reverse next
           k nodes. So move the current pointer
           to skip next k nodes */
        count = 0;
        while (count < k - 1 &&
               current != null)
        {
            current = current.next;
            count++;
        }

        /* 4) Recursively call for the list starting
           from current->next. And make rest of the
           list as next of first node */
        if (current != null)
        {
            current.next =
                    kAltReverse(current.next, k);
        }

        /* 5) prev is new head of the
           input list */
        return prev;
    }

    void printList(Node node)
    {
        while (node != null)
        {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    void push(int newdata)
    {
        Node mynode = new Node(newdata);
        mynode.next = head;
        head = mynode;
    }

    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();

        // Creating the linkedlist
        for (int i = 20; i > 0; i--)
        {
            list.push(i);
        }

        System.out.println("Given Linked List :");
        list.printList(head);
        head = list.kAltReverse(head, 3);
        System.out.println("");
        System.out.println("Modified Linked List :");
        list.printList(head);
    }
}
// This code is contributed by Mayank Jaiswal

Output:

Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19

Time Complexity: O(n)

Method 2 (Process k nodes and recursively call for rest of the list): 
The method 1 reverses the first k node and then moves the pointer to k nodes ahead. So method 1 uses two while loops and processes 2k nodes in one recursive call. 

This method processes only k nodes in a recursive call. It uses a third bool parameter b which decides whether to reverse the k elements or simply move the pointer.  

_kAltReverse(struct node *head, int k, bool b)
  1)  If b is true, then reverse first k nodes.
  2)  If b is false, then move the pointer k nodes ahead.
  3)  Call the kAltReverse() recursively for rest of the n - k nodes and link 
       rest of the modified list with end of first k nodes. 
  4)  Return new head of the list.

Java

// Java program to reverse alternate
// k nodes in a linked list
class LinkedList
{
    static Node head;

    class Node
    {
        int data;
        Node next;

        Node(int d)
        {
            data = d;
            next = null;
        }
    }

    /* Alternatively reverses the given
       linked list in groups of given
       size k. */
    Node kAltReverse(Node head, int k)
    {
        return _kAltReverse(head, k, true);
    }

    /* Helper function for kAltReverse().
       It reverses k nodes of the list only
       if the third parameter b is passed
       as true, otherwise moves the pointer k
       nodes ahead and recursively calls itself  */
    Node _kAltReverse(Node node,
                      int k, boolean b)
    {
        if (node == null)
        {
            return null;
        }

        int count = 1;
        Node prev = null;
        Node current = node;
        Node next = null;

        /* The loop serves two purposes
           1) If b is true, then it reverses
           the k nodes
           2) If b is false, then it moves
           the current pointer */
        while (current != null && count <= k)
        {
            next = current.next;

            /* Reverse the nodes only
               if b is true*/
            if (b == true)
            {
                current.next = prev;
            }

            prev = current;
            current = next;
            count++;
        }

        /* 3) If b is true, then node is the
           kth node. So attach the rest of
           the list after the node.
           4) After attaching, return the new
           head */
        if (b == true)
        {
            node.next =
                 _kAltReverse(current, k, !b);
            return prev;
        }

        /* If b is not true, then attach rest
           of the list after prev. So attach rest
           of the list after prev */
        else
        {
            prev.next = _kAltReverse(current, k, !b);
            return node;
        }
    }

    void printList(Node node)
    {
        while (node != null)
        {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    void push(int newdata)
    {
        Node mynode = new Node(newdata);
        mynode.next = head;
        head = mynode;
    }

    // Driver code
    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();

        // Creating the linkedlist
        for (int i = 20; i > 0; i--)
        {
            list.push(i);
        }
        System.out.println("Given Linked List :");
        list.printList(head);
        head = list.kAltReverse(head, 3);
        System.out.println("");
        System.out.println("Modified Linked List :");
        list.printList(head);
    }
}
// This code is contributed by Mayank Jaiswal

Output:

Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19

Time Complexity: O(n) 

Auxiliary Space: O(n) For call stack because it is using recursion

Please refer complete article on Reverse alternate K nodes in a Singly Linked List for more details!