Solution of Coding challenge #2 by @armandocat

I have participated in (Coding challenge #2)(https://steemit.com/contest/@armandocat/coding-challenge-2-win-sbd-if-you-solve-the-problem) organized by @armandocat and here is my solution.

The Coding Challenge Was

In a Linked list, each node has a value and a next node. Given a Linked list:

Write a function that reverse the linked list, using a recursive algorithm. Write a function that reverse the linked list, using a iterative algorithm. For instance: given A -> B -> C it returns C -> B ->A You don't have to clone the nodes, you must just work on their next properties.

Since the Preferred Language was JavaScript, I have written in that. There is a starting code which you can get it from here https://pastebin.com/tHBtHZVF.

The solution for Iterative Reversal of Linked List. Iterative Reversal means you can reverse each element one by one by using a loop. I have used while loop.

function reverseIterative(obj) {
  // Keeping the copy of obj locally and making original object as null
  var node = obj;
  obj = null;

  // Starting the while loop
  while(node) {
   // In Javascript you can get the next element in the linked list using .next
   // So keeping the next element as a local copy
    var nextNode = node.next;
   // In the next three lines we are just swapping the elements from Previous to Next
    node.next = obj;
    obj = node;
    node = nextNode;
  }
  // Once we get our object, send the object back
  return obj;
}

The solution for Recursive Reversal of Linked List. Recursive meaning that the function will call itself.

function reverseRecursive(obj) {
  // We will check if the obj or obj.next is null or undefined, if it is just return it
  if (!obj || !obj.next) {
    return obj;
  }
 // Keeping a local copy of obj
  var nextNode = obj.next;
  // Calling the Recursive Function using the next node (Recursion)
  var currentNode = reverseRecursive(nextNode);
  //Now the resursion function will be called until the obj is not null.
  obj.next = null;
  nextNode.next = obj;
  // For each funtion call we will return the currentNode
  return currentNode;
}

The output is shown below :

You can get the full solution here: https://repl.it/LWfO/0. A lot of improvements can be made, like removing the unnecessary local variable. Any suggestion is welcomed.

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