Reverse a linked list in JavaScript

Today’s leetcode problem involved linked lists again. This time the task was to reverse one.

I honestly had no idea how to do this and had to look it up. Below is the code I ended up with, and an explanation.

var reverseList = function (head) {
  let current = head;
  let previous = null; //because when we start there is no previous
  let next = null;

  //while current node is not null
  while (current) {
    //keep track of what our next node is
    next = current.next;

    //first time through, the new end of our list is null
    current.next = previous;

    //our new previous is the one we just looked at
    previous = current;

    //move on to the next node
    current = next;
  }

  //previous is now new head
  return previous;
};

To help me explain, here is a simple diagram of a linked list.

This code presents an iterative approach to the problem. You could also solve it recursively.

For this method, you iterate over each node using a while loop, and change the pointers for each node.

Initialize the variables

To use the while loop, you need to set a current variable. The loop will run while this is not null, or until we reach the end of the list, which will have no next pointer.

We also need to keep track of the previous node in the list, as well as the next one. previous is set to null at the beginning, since we’re on the first node, and we’ll deal with the next value in the loop

Setting next and resetting current’s pointer

It’s crucial to save the next variable first thing during a new iteration. That’s because the following step replaces the pointer for current, so we lost whatever next used to be.

Let’s keep track of our variables for the linked list diagram above.

next now equals node 2.

Then, update current’s pointer to previous. current.next now equals null.

Resetting the variables for the next iteration

For our next iteration, previous will be whatever node we just updated in this iteration.

So, previous now equals the current node, which in this case is node 1

The current node for our next iteration is what we saved earlier in our next variable, which is 2

End of list

Once we reach the end, we’ll have previous equal to our last node, 3. This is the new head. The pointers have all been reversed, so we return the new head node. It’s the start of the reversed linked list.