How can I check if two lists are equals?

Question

I have to solve one problem, I don't know the reason why my code doesn't work. I have to check if two lists I created are completely equals so they have the same value at the same position. I'm allowed to use loops as well, even by I prefer the recursive mode. Thank you so much for your help and time!

public static boolean checkEquality(Node n, Node m) {
        if(n != null && m != null) {
            boolean res = false;
            while(n!=null) {
                if(n.getElem()==m.getElem()) {
                    n = n.getNext();
                    m = m.getNext();
                    res = true;
                }
                else
                {
                    res = false;
                }
            }
            return res;
        }
        else
        {
            System.out.println("Lists empty!");
            return true;
        }
    }

Well for starters, consider the case where one list is empty and one is not. You consider those to be equal at the moment. — Michael, Dec 07 '22 at 16:28
So you never call `m.getNext()`? You only ever compare against the first element? — Siguza, Dec 07 '22 at 16:31
You should `break` as soon as you find 2 that aren't equal. Otherwise, this will say they match even if all but the last are unequal. (But there is still the issue you only call `n = n.getNext();` when 2 are equal). — 001, Dec 07 '22 at 16:32
Ok I put as well: m = m.getNext(); But the problem is that it doesnt even enter inside the if — Shukur Nuriyev, Dec 07 '22 at 16:35
a) The calls to `next` should be outside of the `if/else`. b) [What is the difference between == and equals() in Java?](https://stackoverflow.com/a/7520464) — 001, Dec 07 '22 at 16:41
Just debug your code, SO is not a debugging service. Just step through line by line. If you don't know how - add println statements. — rzwitserloot, Dec 07 '22 at 16:41
HINT for the others looking at this: on a mismatch, the pointers aren't advanced, so it hangs forever. The most direct of a few problems here. — rzwitserloot, Dec 07 '22 at 16:42

score 1 · Answer 1 · answered Dec 07 '22 at 16:56

1

There are a couple of weak spots, so I give the solid solution:

public static boolean checkEquality(Node n, Node m) {
    while (n != null && m != null) {
        //if (!Objects.equals(n.getElem(), m.getElem())) {
        if (n.getElem() != m.getElem()) {
            return false;
        }
        n = n.getNext();
        m = m.getNext();
    }
    return n == null && m == null;
}

Comparing can only be done while both n and m are not null. Your code only checks n.
== is not valid for instance for String. Instead of .equals one might also use Objects.equals which also tests for null.
getNext in every loop step.
two empty lists are also equal. Both lists should end at the same time.

The tst fails as soon as two compared nodes are not equal. So one should start with assuming a true result. And as soon as the comparison fails, one should no longer loop and certainly not overwrite res from false to true.

answered Dec 07 '22 at 16:56

Joop Eggen

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The problem I saw on my code by printing is that it doesnt even enter in the if.. I do not understand whats wrong.. – Shukur Nuriyev Dec 07 '22 at 18:19
What are the elements? If `int` or some primitive type then `==` is correct. If `String` or some other class then `x.equals(y)` is needed.. – Joop Eggen Dec 07 '22 at 19:02
I did it, it literally does not enter into the if. It's like it doesn't read it. – Shukur Nuriyev Dec 07 '22 at 19:20
Then you need to debug it, or do `System.out.println(n.getElem());` and such. – Joop Eggen Dec 08 '22 at 08:13

Sayyid Iskandar Khan · Answer 2 · 2022-12-07T17:00:03.503

it would help if you elaborate what type of list u are comparing, linkedlist or arrays. based on your function, it seems that you are planning to compare a linkedlist.

   // sample comparison
   boolean areIdentical(Node a_head, Node b_head) {
        Node a = a_head, b = b_head;
        while (a != null && b != null) {
            if (a.data != b.data)
                return false;
 
            /* If we reach here, then a and b are not null
               and their data is same, so move to next nodes
               in both lists */
            a = a.next;
            b = b.next;
        }
 
        // If linked lists are identical, then 'a' and 'b'
        // must be null at this point.
        return (a == null && b == null);
    }

How can I check if two lists are equals?

2 Answers2