Verifying if two numbers are equal using ArrayLists

Question

I have some question about the code below. It works correctly, but:

• Is there a better/another way to solve the exercise (as an expert would have done:) or mine it's perfectly done?
• Also, I don't see the point of using ArrayList instead of a normal vector like int a[].
• In the method analysis I wanted to set as parameters (ArrayList a,ArrayList b) but an error (incompatible types List cannot be converted to ArrayList) appears in this int a = Testing.analysis(seq1, seq2, t1); line. I thought I was working with ArrayList since I instantiated my sequences as such.

Exercise:

Perform a program that uses a function that returns a 1 if two very large numbers (digits greater than 5 and where each digit is an element of the arrayList) are equal and a 0 otherwise.

    Example 1:

    Entry

    Number 1: 2 3 4 5 6 7

    Number 2: 1 4 6 8 9 0

    Output "They are not equal numbers"

---

class Testing { static int analysis(List a, List b, int t) throws Exception { //using //ArrayList instead of List marks an error, why? int c = 0; if (a.size() == b.size() && a.size() > 5) { for (int i = 0; i < t; i++) { if (a.get(i) == b.get(i)) { c++; } } } else { System.out.println("oops! the lengths are not the same or some of the sequence is less than 6"); throw new Exception(); //to finish the method } if (c == t) { return 1; } else { return 0; } } } public class Tarea24agoENGLISH { public static void main(String[] args) throws Exception { Scanner q = new Scanner(System.in); List seq1 = new ArrayList(); //I wanted to write this List <int> seq=new ArrayList<int>(); but it marked an error, why? System.out.println("Introduce the length of the sequence1 "); int t1 = q.nextInt(); for (int i = 0; i < t1; i++) { System.out.println("Introduce the digits for sequence1"); int n = q.nextInt(); seq1.add(n); } List seq2 = new ArrayList(); System.out.println("Introduce the length of the sequence2"); int t2 = q.nextInt(); for (int i = 0; i < t2; i++) { System.out.println("Introdce the digits for sequence2"); int n = q.nextInt(); seq2.add(n); } int a = Testing.analysis(seq1, seq2, t1); if (a == 1) { System.out.println("same numbers"); } else { System.out.println("\n different numbers"); } } } 

Answer 1

Declaring a method

static int analysis(List a, List b, int t) throws Exception { //using 

This could be just

 public static int isEqual(List<Integer> a, List<Integer> b) { 

This should not be an ArrayList, as that is an implementation. You want to use the interface here because that accepts more things without adding limitations.

You have to say <Integer> rather than <int> because generics have to be objects, not primitives.

If you have more questions about why variants of your code are getting errors, first try searching on Stack Overflow. If you can't find anything, then post a MCVE on Stack Overflow. As was done here.

In general, you want to give methods verb names, like analyze or isEqual. I find isEqual more descriptive.

You don't need to throw a checked Exception. You could throw an unchecked exception like RunTimeException or IllegalArgumentException. If you throw an unchecked exception, you don't need the throws declaration.

You don't need t as it is just a.size(). So just use a.size().

Handling exceptional cases

 if (a.size() == b.size() && a.size() > 5) { 

Consider flipping the logic around.

 if (a.size() < 6) { throw new IllegalArgumentException("The numbers have to have at least 6 digits."); } 

Now code and message match. We don't have to figure out that > 5 means 6 or more. And then negate that for the else. We don't need an else because throw ends the execution of the method. Implicitly, the rest of the method is in an else.

 if (a.size() != b.size()) { return 0; } 

You could throw an exception here, but as previously mentioned, this feels more like a false condition. So you can return your false value.

Early return

 for (int i = 0; i < t; i++) { if (a.get(i) == b.get(i)) { c++; } } 

This could be

 for (int i = 0; i < a.size(); i++) { if (a.get(i) != b.get(i)) { return 0; } } 

If you find an unequal value, you can return immediately.

 if (c == t) { return 1; } else { return 0; } 

And this could be just

 return 1; 

By returning early, you actually simplify the code.

Input to a list

 System.out.println("Introduce the length of the sequence1 "); int t1 = q.nextInt(); 

This is a weird way to do it. If you do it this way, you can just use an array rather than a List. Normally when you use a List, you would get digits until something happened. For example, something like:

 System.out.println("Introduce the digits for sequence1"); for (char digit : q.nextLine().toCharArray()) { seq1.add(digit - '0'); } 

Subtracting '0' from digit converts digit from a character to a number. You could also do Character.digit(digit, 10) which is arguably more portable and flexible.

You may have to cast to Integer if there is a warning.

Anyway, this allows you to enter the digits all next to each other and terminate the number by hitting return.

Answer 2

First thing I noticed don't count how many digits are equal, soon as 1 digit is unequal return false, if the loop finishes return true. Also it seems to me if the sizes are unequal the proper response would be false not an error.