CodeGym/Hilfe bei Java-Aufgaben/whats wrong?

Level 10

Göttingen

15.06.2020
744Aufrufe
12Kommentare

whats wrong?

Frage zur Aufgabe

Java Syntax, Level 7, Lektion 9

Gelöst

whats wrong?

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<String>(); 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) { Iterator<String> a= liste.iterator(); while(a.hasNext()){ String b =a.next(); if(b.liste.contains("r")&&b.liste.contains("l")||b.liste.contains("R")&&b.liste.contains("l")||b.liste.contains("r")&&b.liste.contains("L")||b.liste.contains("R")&&b.liste.contains("L")){ } if(b.liste.contains("r")||b.liste.contains("R")){ liste.remove(); } else if(b.liste.contains("l")||b.liste.contains("L")){ liste.add(); } } return liste; } }

Kommentare (12)

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Level 41 , Bayreuth, Germany

15 Juni 2020, 19:22

problem is, the first if. If you have a r and a l then you do nothing but proceed to the next if clause (if r). And of course a r can be found... try adding a continue; in the body of your first if or make an else if out of your second.

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Level 10 , Göttingen, Germany

16 Juni 2020, 06:11

seems like its not the only poblem .compiler says it can`t find the Iterator but I dont see why?

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Level 41 , Bayreuth, Germany

16 Juni 2020, 07:28

OK, some more hints b.liste.contains("r") is a somewhat wrong syntax. The string you compare is the iterator.next(), therefore b. So it should look like b.contains(... liste.remove() needs an index or you use remove() on the iterator liste.add() needs the object you like to add and an index (in this task) where to add it. Maybe you want to have some index variable, too. Or experiment with a.nextIndex() and let me know how it works :)

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Level 10 , Göttingen, Germany

16 Juni 2020, 13:21

I tried some of your suggestions but I am still getting the same error.I trying it now with a for loop again.

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Level 41 , Bayreuth, Germany

16 Juni 2020, 17:08

If you don't get it going... just answer here with what you have (in code tags ;) )

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Level 10 , Göttingen, Germany

16 Juni 2020, 18:05

public class Solution { public static void main(String[] args) throws Exception { ArrayList<String> liste = new ArrayList<String>(); 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(String b : liste){ if(b.contains("r")&&b.contains("l")||b.contains("R")&&b.contains("l")||b.contains("r")&&b.contains("L")||b.contains("R")&&b.contains("L")){ } else if(b.contains("r")||b.contains("R")){ liste.remove(b); } else if(b.contains("l")||b.contains("L")){ liste.add(b); } } return liste; } }

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Level 41 , Bayreuth, Germany

17 Juni 2020, 07:20

here's a solution that I have written with an iterator.. I used ListIterator over Iterator cause of its add method. Otherwise I would have had to use a variable to keep track of the index. Then using an iterator is kind of useless. Btw., I didn't take care of upper- and lowercase as this isn't asked in the englisch version and the german validation doesn't check that as well. So you may have to test this with lowercase test words or add the checks for uppercase.

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 doesn't contain a r and a l, if so continue the loop
     if (!actualIteration.contains("r") && !actualIteration.contains("l"))
         continue;
     //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;

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Level 41 , Bayreuth, Germany

17 Juni 2020, 07:37

And here my original solution using a index variable (english version therefore list instead of liste) In the for loop we count till list.size(). Keep in mind, that list.size() just gets the size once when initializing the loop. If we later modify the list, it does't keep track of that and may throw an exception. Therefore we need to additionally increase or decrease the index when adding or removing elements from the list. I have to say I like the iterator version more ;)

for (int i = 0; i < list.size(); i++) {
    //if no r and no l or l and r -> continue
    if (!list.get(i).contains("r") && !list.get(i).contains("l"))
        continue;
    if (list.get(i).contains("r") && list.get(i).contains("l"))
        continue;

    if (list.get(i).contains("r")) {
        list.remove(i);
        //decrement index
        i--;
    } else {
        list.add(i, list.get(i));
        //skip the added element by an additional incrementation of the index
        i++;
    }

}
return list;

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Level 10 , Göttingen, Germany

17 Juni 2020, 07:48

thank you verry much for your efforts. But somehow I still get the same error as the countless times before. I even did reset the task several times and pasted in both of your Solutions. for the first Solution i did import the listIterator and then I would get an Output which would be Rose , and liebe. And if added an e to list in the seccond solution same result : no error but the same false Output.

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Level 41 , Bayreuth, Germany

17 Juni 2020, 07:57

Lösung

as said above, it doesn't check for uppercase as this is not asked (validation doesn't check it) but you may add it. Or you change your test words to rose, liebe, leier Adding support for uppercase letters would mean to check if (r && l || R && L || R && l || r && L)

+2

Level 10 , Göttingen, Germany

17 Juni 2020, 09:08

thanks again. it worked now.

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hidden #10655880

Level 9

8 August 2020, 18:54

i have a question about the contains() method, i hope you dont mind me asking :P When the method is called, does it go through every letter of the string, to see if theres a "r" or "l" ? my initial thought when seeing the task was to do a nested for loop (since its a list of string arrays), but then i saw the solution above which looks cleaner to me. i hope you have an answer :D

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