This community-built FAQ covers the “Reverse a Singly-Linked List” code challenge in JavaScript. You can find that challenge here, or pick any challenge you like from our list.
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Hello everyone, 
I resolved that challenge and want to share my answer with you :
function reverseLinkedList(linkedList) {
//linkedList argument represents the head of the Linked List
let prevPointer = null
let currentPointer = linkedList
let nextPointer = null
while (currentPointer !== null) {
//the pointer that should come after the current pointer now comes before the current pointer
prevPointer = currentPointer.next
//the first pointer will be the last so the next pointer of the current pointer should be null for the first iteration
currentPointer.next = nextPointer
//the pointer that will be the next pointer must be the current pointer
nextPointer = currentPointer
//the current pointer should now be the previous pointer of the initial current pointer
currentPointer = prevPointer
}
return nextPointer
}
Enjoy coding everyone ! 
4 Likes
I was a bit lazy and just put all the nodes in the list into an array,
and then used the indices in the array to switch the .next
function reverseLinkedList(linkedList) {
let nodeList = [];
let current = linkedList;
while(current) {
nodeList.push(current);
current = current.next;
}
const lastIndex = nodeList.length - 1;
for (let i = 0; i < lastIndex; i++) {
nodeList[i + 1].next = nodeList[i];
}
if (lastIndex >= 0) {
nodeList[0].next = null;
return nodeList[lastIndex];
}
}
Hey guys,
I proudly present my code solution to the reverse linked list challenge.
function reverseLinkedList(linkedList){
let currentNode = linkedList;
let previousNode = null;
let newNode = null;
while(currentNode !== null) {
newNode = new Node(currentNode.data)
newNode.next = previousNode;
previousNode = newNode;
currentNode = currentNode.next;
}
return newNode;
}
I appear to have been ever lazier 
function reverseLinkedList(linkedList){
const listArray = [];
let current = linkedList;
while (current) {
listArray.push(current.data);
current = current.next;
}
listArray.reverse()
return makeLinkedList(listArray);
}
1 Like
Finally solved it, using the first function.
Had to look it up, not being able to fix it at first.
First i wrote the second function, that worked but didn’t pass.
Tried to keep the runtime shorter by implementing edge cases.
Do you think the edge cases are helping? Or neglectable?
Such a simple elegant solution, for what seems like a difficult problem;
PS: the code below will throw an error, based on the missing imports
const makeLinkedList = require('./makeLinkedList.js');
// O(n) time & O(1) space
function reverseLinkedList(head){
let prev = null;
let next = null;
// we are going to reverse the arrows /pointers
//the first node will become the last, and will therefore point to null
// we will do this by storing the head's pointer first in a a varibale.pointing it to the previous node'.
//first we have to store the head's next node in a a variable, to save it for the next iteration
while (head !== null){
//first we have to store the head's next node in a a variable, to save it for the next iteration
next = head.next;
//then we reset the head nodes pointer to the previous node. in the first iteration, this will be null
head.next = prev;
//now, we set the prev variable to the head, for the next iteration. This is so in the next iteration, the head.next can be set to the previous node.
prev = head;
//now we continue the iteration cycle, by changing the head variable to the previously stored next node: the next node in the original list for processing.
head = next;
}
//when head > the last node of the list, the loop breaks. we now return prev, since that it is the last processed node of the iteration cycle, therfore making it the new head
return prev;
}
function reverseLinkedListX(linkedList){
// Write your code here
let prev = null;
let head = linkedList;
let next = null;
//check if the list has two or more items
if (head.next === null){
console.log('Only one item in the list, no need to reverse!');
return head;
}
//continue if it has more than one item
next = head.next;
//if it only has two items
if (next.next === null){
next.next = head;
head.next = prev;
return next;
}
else {
// has more then two items
while (head !== null){
next = head.next;
head.next = prev;
prev = head
head = next;
}
return prev;
}
}
//Calling your function on an example
console.log("Original")
let exampleLinkedList = makeLinkedList([3,2,1]);
exampleLinkedList.print()
console.log("Reversed")
let reversed = reverseLinkedList(exampleLinkedList);
reversed.print()
// Leave this here so we can test your code
module.exports = reverseLinkedList;
function reverseLinkedList(linkedList){
// Write your code here
let node = linkedList;
let newHead = null;
while(node != null){
let newNode = {data:node.data, next:null};
newNode.next = newHead;
newHead = newNode;
node = node.next;
}
return newHead;
}
Here’s my code for this challenge. I use two pointers to reverse the direction of each link in O(n) time and return the end (new head) node.
const Node = require('./Node.js');
const makeLinkedList = require('./makeLinkedList.js');
function reverseLinkedList(linkedList){
if (!linkedList) {
return linkedList
}
// Write your code here
let x = linkedList
let y = linkedList.next
x.next = null
while (y) {
const temp = y.next
// Reverse direction
y.next = x
x = y
y = temp
}
return x
}
//Calling your function on an example
console.log("Original")
let exampleLinkedList = makeLinkedList([1, 2, 3])
exampleLinkedList.print()
console.log("Reversed")
let reversed = reverseLinkedList(exampleLinkedList)
reversed.print()
// Leave this here so we can test your code
module.exports = reverseLinkedList;
Hello,this my solution))
Preformatted textconst Node = require(‘./Node.js’);
const makeLinkedList = require(‘./makeLinkedList.js’);
function reverseLinkedList(linkedList){
let prev =null
let current = linkedList
while(current){
const next = current.next
current.next = prev
prev = current
current = next
}
// Write your code here
return prev
}
//Calling your function on an example
console.log(“Original”)
let exampleLinkedList = makeLinkedList([4, 8, 15])
exampleLinkedList.print()
console.log(“Reversed”)
let reversed = reverseLinkedList(exampleLinkedList)
reversed.print()
// Leave this here so we can test your code
module.exports = reverseLinkedList;
I got roughly the same answer as everyone else. I used 3 pointers: variables for previous, current, and next node.
Summary
const Node = require('./Node.js');
const makeLinkedList = require('./makeLinkedList.js');
function reverseLinkedList(linkedList){
if (linkedList == null) {
return linkedList;
}
let originalHead = linkedList;
let previous = null;
let current = originalHead;
let next = current.next;
while(next) {
next = current.next;
current.next = previous;
previous = current;
current = next;
}
const newHead = previous || originalHead;
return newHead;
}
class Node {
constructor(value = null) {
this.value = value;
this.next = null;
};
print() {
let current = this;
let output = "" + current.value;
while (current.next) {
current = current.next;
output += " -> " + current.value;
}
output += " -> None";
console.log(output);
//return output;
}
}
function makeLinkedList(array) {
let head = null;
let current;
let previous = null;
for (let element of array) {
current = new Node(element);
if (previous) {
previous.next = current;
}
else {
head = current;
}
previous = current;
}
return head;
}
function reverseLinkedList(linkedList){
if (linkedList == null) {
return linkedList;
}
let originalHead = linkedList;
let previous = null;
let current = originalHead;
let next = current.next;
while(next) {
next = current.next;
current.next = previous;
previous = current;
current = next;
}
const newHead = previous || originalHead;
return newHead;
}
//Calling your function on an example
console.log("Original");
let exampleLinkedList = makeLinkedList([4, 8, 15])
exampleLinkedList.print();
console.log("Reversed");
let reversed = reverseLinkedList(exampleLinkedList);
reversed.print();
Hello evrybody. it is my solution 
const Node = require(‘./Node.js’);
const makeLinkedList = require(‘./makeLinkedList.js’);
function reverseLinkedList(linkedList){
// Write your code here
if (linkedList === null || linkedList.next === null) {
return linkedList;
}
let reversedHead = reverseLinkedList(linkedList.next);
linkedList.next.next = linkedList;
linkedList.next = null;
return reversedHead;
}
//Calling your function on an example
console.log(“Original”)
let exampleLinkedList = makeLinkedList([4, 8, 15])
exampleLinkedList.print()
console.log(“Reversed”)
let reversed = reverseLinkedList(exampleLinkedList)
reversed.print()
// Leave this here so we can test your code
module.exports = reverseLinkedList;
1 Like
I kindda like your recursive solution
Here is my solution
function reverseLinkedList(head){
let prev = null
let next = null
while ( head ) {
next = head.next // Save next Node
head.next = prev // Assign previous Node as Next Node
if ( !next ) { break } // Break if no next Node
prev = head // Current head becomes previous Node
head = next // Next Node becomes current head
}
return head
}
At first, I came up whit this extremely and unnecessarily complex solution:
function reverseLinkedList(linkedList){
function findNthNode(linkedList, n) {
let count = 0;
let tail = linkedList;
let current = null;
while(tail) {
tail = tail.next;
if(count >= n) {
if(!current) {
current = linkedList;
}
current = current.next;
}
count++;
}
if(!current && count === n) {
return linkedList;
}
return current;
}
let secondToLast = findNthNode(linkedList, 2);
let tail = findNthNode(linkedList, 1);
let current = linkedList;
linkedList = tail;
while(secondToLast) {
tail.next = secondToLast;
secondToLast.next = null;
secondToLast = findNthNode(current, 2);
tail = findNthNode(current, 1);
}
return linkedList;
}
Also, this solution is not at all time-performant, being O(N²).
Then, I cheated a little bit 
since I read it could be solved with O(N) time complexity, and searched for tips on how to achieve this, ending up with this solution (very similar to what others have already posted here):
function reverseLinkedList(linkedList) {
let previousPointer = null;
let currentPointer = linkedList;
while(currentPointer) {
const nextPointer = currentPointer.next;
currentPointer.next = previousPointer;
previousPointer = currentPointer;
currentPointer = nextPointer;
}
return previousPointer;
}
