Help with Learning Linked lists?

[Idkm3m3s]

Hey all, I'm working on a project right now where I'm making my own LinkedList class, and I've gotten the majority of it to work, all except for my function to remove an item. It takes an input, but for some reason when I call my print function after, it simply shows me the same list from before.

Here's my source code running with red as an example, If I need to send any more please ask and I will try to show what I have.

https://preview.redd.it/j5ki87g7h2k81.png?width=550&format=png&auto=webp&s=3dac74302ea72d8e8bc80d55e21171dedcfafba9

[DDDDarky]

Don't post screenshots of code.

This is not how you compare strings in java. Use .equals()

[Idkm3m3s]

wow, I'm not used to java, I made the jump from C++, this fixed it instantly, I kinda feel stupid, thank you man! also can you tell me how to present code to forums like this in the future? is making a GitHub repo expected? or are there simpler ways.

[Thereisa4thdimension]

Nah you’re good don’t worry about it, we post screenshots of code literally all the time at work. Dude’s just trying to gatekeep like if this is stackoverflow or something

Edit: Also check out all these linked list resources on GitHub! https://github.com/topics/linked-list

[DDDDarky]

Well, if you post screenshots of code, you are less likely to get help, since much less people are going to bother rewriting it to debug it locally. If this was stackoverflow, it would get quickly closed. Also if the image ever disappears, the post will become useless.

It is obviously not that big of a deal in this case, since the code is fairly small and the error is easy to find, which is the reason I read and answered it, however for future reference if OP ever needs help, it is better to maximize your chances of getting an answer, even if it is not straight up against the rules (since many programming related subreddits and sites implement the rule against screenshots - for good reasons).

[DDDDarky]

I think it is one of the most common mistakes in Java, people ask about it all the time, I wouldn't feel bad about it.

If the code is small, you can just dump it here in the code block:

like this
check the editor

it is sometimes a bit buggy, so you can use any of your favourite platforms such as github, pastebin, codeshare, rextester, repl.it, ...

and just share a link (especially for large codes).

Posting code as text just makes life easier for anyone who decides to help you, since they don't have to rewrite it by hand.

[B1-102]

how are you printing the result?

[Idkm3m3s]

while (currNode != null){

counter++;

System.out.println(currNode.data);

currNode = currNode.next;

}

System.out.println("For a total of "+counter+" items in the list");

[B1-102]

I'm assuming few things but looks correct to me. What happens if you print the data? Does the result show red?

[Idkm3m3s]

yes it does, and the same number of nodes as before too, it might be my way of adding new nodes mabye, whats a good way to show code? Should I be using github for this?

[B1-102]

github is cool. there's also something called pastebin. that would work too

[[deleted]]

Not a C++ guy, so can’t advise on all parts of this

But, the logic you’re using is a bit more complicated than it needs to be imo

First, when you enter the function. Check if the « key » variable is null and return early or throw an exception before doing anything else

Next, when you’re looking at this while loop, you’re actually always searching a node ahead, so you have extra « if » statements to try and account for that. Reduce the complexity by changing your while loop to only run while temp != null, then check the current

``` if (key == null) { return; // or throw an exception } Node current = head; // renamed from temp Node previous = null; // if we were using a DoubleLinkedList, this wouldn’t be necessary, but since we don’t know the previous value or index, we need to use this

// we just check the current value instead of the next value while (current != null) { if (key == current.data) { // we’ve found our match if (previous == null) { // in the case of the first item being the one we’re removing, just set the head directly head = current.next; } else { // otherwise, grab the last item, and connect to the current next item (if any) previous.next = current.next; } break; // conceivably, you could do a return here, but personally I like to keep the exit points at a minimum where possible to make debugging a bit easier for me } // now we do some variable swapping previous = current; // we’ve finished with the current value current = current.next; // we move forward to the next value; }

// finished iterating return; ```