# 206. Reverse Linked List

## The Problem

Link to original problem on Leetcode

Reverse a singly linked list.

## Example

```
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
```

**Follow up**: A linked list can be reversed either iteratively or recursively. Could you implement both?

## My Solution

Here are my attempts to reverse the list both iteratively and recursively.

### Iterative Version

Here I create variables to store the list I'm iterating through, and the new one I'm building up. I go through the original list with a while loop, setting the `reversedList`

to equal a new node with the `currentNode`

's value as its own and the previous value of `reversedList`

as the next node. Then the `currentNode`

is set to it's next value. The loop terminates when the next node of `currentList`

is `null`

.

Got a 98.12% on speed (68ms), but only 38.39% (38.7mb) on memory.

`/**`

* Definition for singly-linked list.

* function ListNode(val, next) {

* this.val = (val===undefined ? 0 : val)

* this.next = (next===undefined ? null : next)

* }

*/

const reverseList = (head) => {

let currentNode = head;

let reversedList = null;

while (currentNode) {

reversedList = new ListNode(currentNode.val, reversedList);

currentNode = currentNode.next;

}

return reversedList;

};

### Recursive Version

If `head`

is `null`

, then return the value of `reversed`

immediately. Otherwise, return the function with `head`

's next node as the new `head`

and a new `ListNode`

for `reversed`

with `head`

's value as the value and the previous value of `reversed`

as the next node.

Got a 86.52% on speed (76ms), but only 5% (39.5mb) on memory.

`const reverseList = (head, reversed = null) => {`

if (!head) return reversed;

return reverseList(head.next, new ListNode(head.val, reversed));

};

## Best Solution

I need to review more of these solutions. Rather than create new ListNodes, they tend to just store one value or another in a temporary variable and then mutate the originals to flip things around.

Example:

`const reverseList = (head, prev=null) => {`

if(!head) return prev;

let temp = head.next;

head.next = prev;

return reverseList(temp, head);

};

const reverseList = (head) => {

let node = head, reversed = null;

while(node){

let temp = node;

node=node.next;

temp.next = reversed;

reversed = temp;

}

return reversed;

};