Reverse linked list without using any extra space.

Linked list reversal I found very common question asked in many companies. One of the easy way to implement this is, using stack and traversing whole singly linked list and dumping into stack. One list linked list finishes, start poping up and preparing new linked list. Problem gets interesting when we are not suppose to use any extra space like stack or another linked list. Here is the implementation of such. I am using java to implement this and approach is test driven development hence test cases are also attached.
The Node class that represent single node -

package com.adnan.linkedlist;

/**
 * User  : Adnan
 * Email : sendtoadnan@gmail.com
 * Date  : 9/21/13
 * Time  : 12:02 PM
 */
public class Node {

    public Node(int value, Node node){
        this.value = value;
        this.node = node;
    }
    private int value;
    private Node node;

    public int getValue() {
        return value;
    }

    public Node getNode() {
        return node;
    }

    public void setNode(Node node){
        this.node = node;
    }
}

Service class that takes start node as input and reserve it without using extra space.

package com.adnan.linkedlist;

/**
 * User  : Adnan
 * Email : sendtoadnan@gmail.com
 * Date  : 9/21/13
 * Time  : 11:54 AM
 */
public class SinglyLinkedListReversal {

    private static final SinglyLinkedListReversal service 
= new SinglyLinkedListReversal();
    public static SinglyLinkedListReversal getService(){
        return service;
    }



    public Node reverse(Node start){
        if (hasOnlyNodeInLinkedList(start)){
            return start;
        }
        Node firstNode, secondNode, thirdNode;
        firstNode = start;
        secondNode = firstNode.getNode();
        while (secondNode != null ){
            thirdNode = secondNode.getNode();
            secondNode.setNode(firstNode);
            firstNode = secondNode;
            secondNode = thirdNode;
        }
        start.setNode(null);
        return firstNode;
    }

    private boolean hasOnlyNodeInLinkedList(Node start) {
        return start.getNode() == null;
    }


}


  

Another logic of reversal could be a recursive approach. So replace reverse method of above class by below and it should return result in recursive manner.

 public Node reverseRecursively(Node start){
     if (start == null ){
         return null;
     }
     Node lastNode;
     if (start.getNode() == null ){
       lastNode = start;
     }else {
         lastNode = reverseRecursively(start.getNode());
         start.getNode().setNode(start);
         start.setNode(null);
     }
      return lastNode;
   }
  

And The test case that covers above scenario. Please note that you require junit jars. I am using testng.jar; you can use any whatever pleases you..

package com.adnan.linkedlist;

import org.testng.annotations.Test;

import static org.testng.AssertJUnit.assertTrue;

/**
 * User  : Adnan
 * Email : sendtoadnan@gmail.com
 * Date  : 9/21/13
 * Time  : 12:11 PM
 */
public class SinglyLinkedListReversalTest {

    private SinglyLinkedListReversal reversalService = 
SinglyLinkedListReversal.getService();

    @Test
    public void test_reverseSingleElement() throws Exception {
        Node node = new Node(1, null);
        reversalService.reverse(node);
        assertTrue(node.getNode() == null);
        assertTrue(node.getValue() == 1);
    }


    //original - Node1(1) -> Node2(2) -> Node3(3)
    //reverse - Node3(3) -> Node2(2) -> Node1(1)
    @Test
    public void test_reverseThreeElement() throws Exception {
        Node node3 = new Node(3, null);
        Node node2 = new Node(2, node3);
        Node start = new Node(1, node2);


        start = reversalService.reverse(start);
        Node test = start;
        for (int i = 3; i >=1 ; i -- ){
          assertTrue(test.getValue() == i);
            test = test.getNode();
        }


    }

    @Test
    public void test_reverseFourElement() throws Exception {
        Node node4 = new Node(4, null);
        Node node3 = new Node(3, node4);
        Node node2 = new Node(2, node3);
        Node start = new Node(1, node2);


        start = reversalService.reverse(start);
        Node test = start;
        for (int i = 4; i >=1 ; i -- ){
            assertTrue(test.getValue() == i);
            test = test.getNode();
        }
    }

        @Test
        public void test_reverse10Element() throws Exception {
            Node node10 = new Node(10, null);
            Node node9 = new Node(9, node10);
            Node node8 = new Node(8, node9);
            Node node7 = new Node(7, node8);
            Node node6 = new Node(6, node7);
            Node node5 = new Node(5, node6);
            Node node4 = new Node(4, node5);
            Node node3 = new Node(3, node4);
            Node node2 = new Node(2, node3);
            Node start = new Node(1, node2);


            start = reversalService.reverse(start);
            Node test = start;
            for (int i = 10; i >=1 ; i -- ){
                assertTrue(test.getValue() == i);
                test = test.getNode();
            }


    }

    @Test
    public void test_reverseTwoElement() throws Exception {
        Node node2 = new Node(2, null);
        Node start = new Node(1, node2);


        start = reversalService.reverse(start);
        Node test = start;
        for (int i = 2; i >=1 ; i -- ){
            assertTrue(test.getValue() == i);
            test = test.getNode();
        }


    }
}