Does anyone see anything wrong with this code?

```java
/**
* This method reverse2() will reverse all data nodes in this list, WITHOUT
* creating(introducing) new list nodes, by simply re-wiring the next reference in
* the existing list node. For example, list1 = []—>[A]—>[B],
* the reversed list1 will be []-->[B]-->[A],
* after assigning node A to B's next reference and setting A's next to null.
*
*/
public void reverse2() {
if(this.size <= 1)
return;
// The following method call works on a *sublist* without a Dummy Node.
// Namely, we preserved the OLD dummy head node in the reversed list.
this.head.next = reverse(this.head.next, this.head.next.next);
}

/**
* Please implement the helper method below for reverse2().
* @param first, the first node of the list to be reversed.
* @param second, the second node of the list to be reversed.
* @return the new head node of the reversed list.
* Note: you are NOT allowed to create new list node, but have to
* re-wiring the existing nodes by changing their next references.
* Write this method using recursion.
*/
private ListNode reverse(ListNode first, ListNode second) {
ListNode curr = first;
ListNode newFirstNode = null;
ListNode newLastNode = null;
ListNode newHead = new ListNode();
while (curr.next != null) {
newLastNode = curr;
newFirstNode = curr.next;
curr = curr.next.next;
}
newHead.next = newFirstNode;
newLastNode.next = null;
newFirstNode.next = reverse(newFirstNode.next, newFirstNode.next.next);
return newHead; //change this line of code as needed.
}
```

and this ^

I'm honestly not sure what the point of passing the second node is, since (I think) it can be determined from the first node using .next

<@&435155896780324864>

That's all for tonight by the way. I won't spam you guys with a bunch of other methods, it's just these.

@Jacob well it's recursive right, and one node IIRC is a null node, right? null wouldn't have a null.next

ah

you mean at the end?

when the function is originally called, I assume both nodes are fine

but ya I think that makes sense

that at some point second will be null

wait no that shouldn't be an issue

because each Linked List object has a size variable to check for that

wait no that wouldn't work because I'm not actually passing a list

haha yeah that linked list/ tree stuff is tricky, maybe later in the weekend I can focus on it if you're still working on it, hopefully someone else here can help out sooner than that though 😁 <@&387091385075105804> <@&435155896780324864>

Do you think I'm understanding this correctly?

```java
/**
* This method reverse2() will reverse all data nodes in this list, WITHOUT
* creating(introducing) new list nodes, by simply re-wiring the next reference in
* the existing list node. For example, list1 = []—>[A]—>[B],
* the reversed list1 will be []-->[B]-->[A],
* after assigning node A to B's next reference and setting A's next to null.
*
*/
public void reverse2() {
if(this.size <= 1)
return;
// The following method call works on a *sublist* without a Dummy Node.
// Namely, we preserved the OLD dummy head node in the reversed list.
this.head.next = reverse(this.head.next, this.head.next.next);
}

/**
* Please implement the helper method below for reverse2().
* @param first, the first node of the list to be reversed.
* @param second, the second node of the list to be reversed.
* @return the new head node of the reversed list.
* Note: you are NOT allowed to create new list node, but have to
* re-wiring the existing nodes by changing their next references.
* Write this method using recursion.
*/
private ListNode reverse(ListNode first, ListNode second) {
ListNode newHead = new ListNode();
if (second.next != null) {
ListNode curr = first;
ListNode newFirstNode = null;
ListNode newLastNode = null;
while (curr.next != null) {
newLastNode = curr;
newFirstNode = curr.next;
curr = curr.next.next;
}
newHead.next = newFirstNode;
newLastNode.next = null;
newFirstNode.next = reverse(newFirstNode.next, newFirstNode.next.next);
}
else {
newHead.next = second;
second.next = first;
first.next = null;
}
return newHead; //change this line of code as needed.
}
```

@ThisIsChris okay this is what I have right now, do you think this works?

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<@&435155896780324864> Does anyone know how to do a Big O time complexity analysis on a Java function?

@Jacob example function?

```java
public static void two(int n)
{
if(n > 0)
{
System.out.println("n: " +n);
two(n - 1);
two(n - 1);
}
else if (n < 0)
{
two(n + 1);
two(n + 1);
System.out.println(″n: ″ + n);
}
}
```

It's mostly like doing it on paper, you have most of the same assumptions of the Random Access Machine

Do you have any skepticism that assumptions of RAM don't hold?

No, it's just a basic analysis. I need to provide a "logical justification" for my answer. Not sure if that means I have to write a paragraph or there's some kind of equation I need to use.

There's an equation.

Not necessarily expecting an answer to the question from anyone, but maybe something that can get me in the right direction

f(0) = O(1)

f(|n|) = 2*f(|n|-1) + O(1)

😬

so f(n) = O(2^n)

Is that the equation for this particular function?

yep

How would I go about getting that?

So start by looking at the function to understand what is the "base case". The base case is the input that doesn't cause a recursive call

ah, okay
and write the base case as one of those formulas?

the only input that doesn't have a recursive call is n=0

yep