CodeGym/Hilfe mit Java-Aufgaben/Schön wäre es gewesen

Level 9

17.07.2020
515Aufrufe
11Kommentare

Schön wäre es gewesen

Frage zur Aufgabe

Gelöst

Was mache ich falsch? Ich habe so viel ausprobiert. Ich hoffte jetzt eine Möglichkeit zu haben das ganze Wort nach den Buchstaben zu durchsuchen. Das Programm hat die Aufgabe erfüllt und doch funktioniert es nicht. Ich hätte gerne einen Tip. Danke im Voraus für Eure Bemühungen

1. Erstelle eine Liste von Wörtern und fülle sie selbst aus.
2. Die korrigieren-Methode muss:
2.1. alle Wörter aus der Liste entfernen, die den Buchstaben „r“ enthalten.
2.2. alle Wörter duplizieren, die den Buchstaben „l“ enthalten.
2.3. wenn ein Wort sowohl „r“ als auch „l“ enthält, dann machen wir nichts.
2.4. alle anderen Wörter unangetastet lassen.

Zum Beispiel:
Rose Leier Liebe

Ausgabe:
Leier Liebe Liebe

Anforderungen:

Das Programm darf keine Daten von der Tastatur lesen.
Die korrigieren-Methode muss alle Wörter, die den Buchstaben "r" enthalten, aus der Liste entfernen. Es gibt eine Ausnahme: Wörter, die sowohl "r" als auch "l" enthalten, müssen unangetastet bleiben.
Die korrigieren-Methode muss Wörter mit dem Buchstaben "l" duplizieren (eine weitere Instanz dieses Wortes zur Liste hinzufügen). Es gibt eine Ausnahme: Wörter, die sowohl "l" als auch "r" enthalten, dürfen nicht dupliziert werden.
Die korrigieren-Methode muss Wörter, die sowohl "l" als auch "r" enthalten, unangetastet lassen.
Die korrigieren-Methode muss Wörter, die weder "l" noch "r" enthalten, unangetastet lassen.

package de.codegym.task.task07.task0716; import java.util.ArrayList; /* R oder L */ public class Solution { public static void main(String[] args) throws Exception { ArrayList<String> liste = new ArrayList<>(); liste.add("Rose"); // 0 liste.add("Liebe"); // 1 liste.add("Leier"); // 2 liste = korrigieren(liste); for (String s : liste) { System.out.println(s); } } public static ArrayList<String> korrigieren(ArrayList<String> liste) { for(int i = 0; i < liste.size(); i++) { if((liste.get(i).indexOf("R")!=-1 ||liste.get(i).indexOf("r")!=-1)&& (liste.get(i).indexOf("L")!=-1|| liste.get(i).indexOf("l")!=-1)) continue; if(liste.get(i).indexOf("R")!=-1 ||liste.get(i).indexOf("r")!=-1) liste.remove(i); if(liste.get(i).indexOf("L")!=-1 ||liste.get(i).indexOf("l")!=-1) liste.add(i, liste.get(i)); i++; } return liste; } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

17 Juli 2020, 17:58

When removing or adding an item in a List the internal indices of all remaining items down the list will shift either +1 or -1 accordingly. Lets say you have a list with three values random values like so: {0:Item1, 1:item2 ,3:item3} (where the number to the left of the colon is the index) Now lets say that you have a for loop and during first loop cycle when i = 0, you remove the item at the index 0, item1, from that list. This will force the remaining items down the list to shift their indices to the left so the list becomes: {0:Item2, 1:Item3} The loop cycle will conclude and the next loop cycle would start and i would be incremented by 1. This would mean the code now checks the value in index 1, which is item3. This effectively has skipped the value "Item2", which will be problematic if item2 was supposed to be duplicated or removed. This is a bug in the code. It looks like this code tries to adjust the value of i when an item is added to the list so that it does not check the same item the next loop cycle, but line 32 is not inside the block of code for the if statement on line 30. The rules for if statements are that if there are no curly brackets after the 'if' statement then it will only process the very next line of code. If more than 1 line of code is meant to execute then it all needs to be enclosed in a block of code:

if(liste.get(i).indexOf("L")!=-1 ||liste.get(i).indexOf("l")!=-1)
   liste.add(i, liste.get(i)); // this line is the only line that will execute if the statement is true
i++; // this line is outside the if block and executes every loop cycle

You will want to correct this by surrounding the intended code to execute in a block of code.

Max

Level 9 , Germany

17 Juli 2020, 19:43

Thank you. First I surrounded the for loop with the missing brackets. After that I swaped lines 30-32 with lines 27-28. I thought, that will help to keep the size, when the loop to remove an index is performed. I checked the program with several words and it went well. But when I checked the program again with the first three words, the word "Liebe" was not duplicated.

Guadalupe Gagnon

Level 37 , Tampa, United States

17 Juli 2020, 19:54

Because "Rose" was removed when i = 0 and "Liebe" was moved to index 0; the loop moved to the second iteration where i = 1, and "Liebe" was skipped.

Max

Level 9 , Germany

17 Juli 2020, 20:57

So, how do I remove Rose but keep the other strings on their origin location? Or do I think in a wrong way? Would it help to put a i++ after the removal?

Guadalupe Gagnon

Level 37 , Tampa, United States

17 Juli 2020, 21:12

"Would it help to put a i++ after the removal?" You are thinking on the correct track, but putting i++ after removal would then make it skip 2 values in the list instead of only 1.

Max

Level 9 , Germany

18 Juli 2020, 05:44

Good Morning. I've got the feeling, that my brain got stuck. If there was a reset button, I would use that. It may sound stupid, because you are already on level 37, but I struggle on this. Now, I was thinking about, what will happen, if I put i++ before the removal. But doesn't extend the ArrayList at the end? So, would that make sense than? Is there another way to keep the rest on their positions? I will try now and see, what happens. Thank you very much for your help. update: I think I got it. Instead of i++, I put I--. Because, if I remove one index, and the rest moves into that empty place, than I have to tell the counter to go back one position to let the loop run from that "removed" position again.

Guadalupe Gagnon

Level 37 , Tampa, United States

18 Juli 2020, 05:53

"update: I think I got it. Instead of i++, I put I--." YES!!! This is all part of the learning process of programming. I went through the same exact thinking as you are right now. It takes time to learn how to really think in computer logic. Come back in a year and look at these tasks and you will be scratching your head on how they were even complicated to begin with.

Max

Level 9 , Germany

18 Juli 2020, 06:04

Thank you again for your help. By the way, isn't it night time in Tampa(Florida?) ? Do you guys there don't sleep? :)))

Nouser

Level 36 , Germany

18 Juli 2020, 07:19

As an alternative I suggest to hava a look at the ListIterator class. It is a beefed up Iterator for lists, that allows adding elements at the current iterator position. Solving the task that way needed (for me) less time than this attempt.

public static ArrayList<String> korrigieren(ArrayList<String> liste) {
    //using a ListIterator over an Iterator cause of the possibility to add entries over the iterator
    //meaning at that position where the iterator is atm
    //using a regular iterator you'd have to use a index variable to determine the actual positon
    ListIterator<String> it = liste.listIterator();
    //check if there is another element in the list
    while (it.hasNext()) {
        //save the actual element in a variable
        String actualIteration = it.next();
        //check if the actual element contains a r and a l, if so continue the loop
        //of course we could add that check to the above one
        if (actualIteration.contains("r") && actualIteration.contains("l"))
            continue;
        //if the actual element contains a r, remove the actual element
        if (actualIteration.contains("r"))
            it.remove();
            //add an element... from the docs:
            //The element is inserted immediately before the element that would be returned by next()
            //so calling next() will skip the inserted element and we don't have to take care of the index
            //however ListIterator could get the actual position as well with nextIndex()
        else if (actualIteration.contains("l")) {
            it.add(actualIteration);
        }
    }
    return liste;
}

Guadalupe Gagnon

Level 37 , Tampa, United States

19 Juli 2020, 20:31

It was night time, but on a Friday that is typical that I am still up.

Max

Level 9 , Germany

20 Juli 2020, 13:55

I see. well, enjoy. take care