The other answer nicely describes better ways to validate dates.

IMHO it fails to address your concern: why does IntelliJ IDEA report "'return' is unnecessary as the last statement in a 'void' method." on some of your return statements?

To explain that, I simplified your example to this:

public static void validar(int n) {
    System.out.println(n);

    if (n == 3) {
        System.out.println("a");
        return;  // Note 1
    } else if (n == 5) {
        System.out.println("b");
        return;  // Note 1
    }

    int x = n * n;

    if (x == 4) {
        System.out.println("c");
        return;  // Note 2
    } else if (x == 16) {
        System.out.println("c");
        return;  // Note 2
    }

}

I've added two notices to the return statements:

Note 1: the return; statements at these places is needed to exit early from the method. Without these return; statements the rest of the method would also execute. Accordingly there is no report from IntelliJ IDEA at these

Note 2: there are no more statements that could possibly be executed after these return; statements. They are therefore unnecessary. They are the last statements in the branches of an if - else if chain and there are no more statements after that chain. That means that execution of the method ends after executing the preceding statement anyway. For these return; statements IntelliJ IDEA reports "'return' is unnecessary as the last statement in a 'void' method.".

Please note that this is an informational message: its intention is to help you write better (cleaner) code. It doesn't mean that your code is faulty (i.e. it doesn't mean your code will not compile or will have bugs in it.)

You can change that simplified code (without changing its observable behaviour) to

public static void validar(int n) {
    System.out.println(n);

    if (n == 3) {
        System.out.println("a");
        return;  // Note 1
    } else if (n == 5) {
        System.out.println("b");
        return;  // Note 1
    }

    int x = n * n;

    if (x == 4) {
        System.out.println("c");
    } else if (x == 16) {
        System.out.println("c");
    }

}

It looks like you are trying to split a string that contains a date. Possibly something like this. 2023-05-11 12:25:32.

Rather than trying to reinvent the wheel, check out some of these other options.

Option 1: SimpleDateFormat

Why not use SimpleDateFormat to do the job for you? This method has been around since the beginning and does the job pretty well. SimpleDateFormat will attempt to parse the date based on a pattern you define. If it fails to parse, it will throw a checked exception.

public static void validar(String data)
{
  DateFormat format = new SimpleDateFormat("yyyy-MM-dd h:mm:ss");
  try 
  {
    format.parse(data);
  } 
  catch (ParseException e) 
  {
    System.out.println(e.getMessage());
    System.out.println("El format de la data és incorrecte. Comproveu que la data coincideixi amb el format correcte " +
      "de \"AAAA-MM-DD hh:mm:ss\"");
  }
}

This way you don't need to worry about all of the complicated logic. Instead, you try to parse the date, and if your parse fails, then the date fails validation.

Option 2: Simple Regex

If you aren't concerned with the numbers actually corresponding to a valid date, and just that they match the format you expect, than you can accomplish this via simple pattern matching.

  public static void validarRegex(String data)
  {
    Pattern pattern = Pattern.compile("^\\d{4}\\-\\d{2}\\-\\d{2} \\d{2}:\\d{2}:\\d{2}$");
    if (!data.matches(pattern.pattern())) {
      System.out.println(data);
      System.out.println("El format de la data és incorrecte. Comproveu que la data coincideixi amb el format " +
        "correcte " + "de \"AAAA-MM-DD hh:mm:ss\"");
    }
  }

Option 3: DateTimeFormatter

As user Ole V.V. pointed out, you can also use the updated DateTimeFormatter to parse the string as well. A word of warning here. The DateTimeFormatter throws an unchecked exception, so you should be prepared for this behavior. My example below does not catch the exception. You can catch it here, or higher up the chain if you choose.

  public static void validar(String data)
  {
    System.out.println(data);
    LocalDateTime.parse(data, DateTimeFormatter.ofPattern("uuuu-MM-dd HH:mm:ss"));
  }

Option 4: Complex Regex

I don't recommend this, because it's nearly impossible to get a good check due to things like leap years. However, if you want to give it a go, there are plenty of examples floating around.

Option 5: String parsing

This is essentially what you are already doing, but a bit more concise.

public static void validate(String data)
{
  System.out.println(data);

  if (isValidDate(data)) {
    String[] arrDiaHora = data.split(" ");
    String diamesany = arrDiaHora[0];
    String horaminutsegon = arrDiaHora[1];

    String[] arrDiaMesAny = diamesany.split("-");
    String strany = arrDiaMesAny[0];
    String strmes = arrDiaMesAny[1];
    String strdia = arrDiaMesAny[2];

    String[] arrHoraMinutSegon = horaminutsegon.split(":");
    String strhora = arrHoraMinutSegon[0];
    String strminut = arrHoraMinutSegon[1];
    String strsegon = arrHoraMinutSegon[2];

    if (arrDiaHora.length != 2) {
      System.out.println("No hi ha dos blocs formats per data i temps");
      System.out.println("Data incorrecta" + "\n");
    }
  }
}

public static boolean isValidDate(String data)
{
  boolean valid = true;
  if (!data.contains(" ")) {
    invalidData("No hi ha separació entre data i temps");
    valid = false;
  }
  else if (!data.contains("-")) {
    invalidData("La data no conté un guions");
    valid = false;
  }
  else if (!data.contains(":")) {
    invalidData("El temps de la data no conté :'s");
    valid = false;
  }
  return valid;
}

public static void invalidData(String message)
{
  System.out.println(message);
  System.out.println("Data incorrecta" + "\n");
}