Can anyone give me a hint?

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Rozwiązane

input: output: rosa luz luz loa loa lira lira loa Can someone tell me, why second condition is failed, when "rosa" disappeared, but third is correct, when "luz" wasn't doubled? What im doing wrong?

1. Utwórz listę wyrazów i sam ją zapełnij.
2. Metoda napraw powinna:
2.1. usuwać z listy wszystkie wyrazy zawierające literę „r”.
2.2. duplikować wszystkie wyrazy zawierające literę "l".
2.3. pozostawiać niezmienione wszystkie wyrazy zawierające zarówno literę "r, jak i "l".
2.4. nic nie robić z pozostałymi wyrazami.

Na przykład:
rosa lira luz

Wyjście:
lira luz luz

Wymagania:

Program nie może wczytywać danych z klawiatury.
Metoda napraw powinna usuwać z listy wszystkie wyrazy zawierające literę „r”. Jest jednak wyjątek: wyrazy zawierające zarówno literę „r”, jak i „l” powinny pozostać niezmienione.
Metoda napraw musi duplikować wyrazy zawierające literę „l” (dodaj inną instancję tego słowa do listy). Jest jednak wyjątek: wyrazy zawierające zarówno literę „r”, jak i „l” nie powinny być duplikowane.
Metoda napraw nie powinna nic robić z wyrazami zawierającymi zarówno „l”, jak i „r”.
Metoda napraw nie powinna nic robić z wyrazami, które nie zawierają ani „l”, ani „r”.

package pl.codegym.task.task07.task0716; import java.util.ArrayList; /* R czy L */ public class Solution { public static void main(String[] args) throws Exception { ArrayList<String> lista = new ArrayList<String>(); lista.add("rosa"); // 0 lista.add("luz"); // 1 lista.add("loa"); // 1 lista.add("lira"); // 2 lista = napraw(lista); // for(int f = 0; f <lista.get(0).length(); f++) { // System.out.println(lista.get(0).charAt(f)); // } for (String s : lista) { System.out.println(s); } } public static ArrayList<String> napraw(ArrayList<String> lista) { // tutaj wpisz swój kod int a = 0; int b = 0; int c = lista.size(); for(int i = 0; i < c; i++) { for(int f = 0; f <lista.get(i).length(); f++) { if (lista.get(i).charAt(f) == 'r') { a++; } else if (lista.get(i).charAt(f) == 'l') { b++; } } if (a != 0 && b != 0) { } else if (a > 0) { lista.remove(i); c--; } else if (b > 0) { lista.add(lista.get(i)); } else { } a = 0; b = 0; } return lista; } }

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Nixodem

Poziom 11 , Poland

15 kwietnia 2022, 05:18

Thanks everyone for Your commitment and comments, it works now! :) Happy Easter everyone

NeedXMafia Backend Developer

14 kwietnia 2022, 13:40

nie wiem co kolega ci mowi ale twoj blad polega ta tym ze usuwasz elementy z listy podczas gdy działasz w pentli na zasadzie listy ktora modyfikujesz w trakcie pentli. działaj na kopi listy a nie orginale i zwroc kopie. dzieje sie tak poniewaz gdy usuwaz element z listy to nastepny przeskakuje na index poprzedniego dzienki czemu jest pomijany W skrócie utworz kopie listy ale sprawdzaj na orginale natomiast usuwaj i dodawaj na kpi oraz zwroc kopie ENG I don't know what your colleague is telling you but your mistake is that you remove items from the list while you are operating in the pentas as a list that you modify during the pent. affect the copy of the list, not the original, and return the copy. This is because when I remove an item from the list, the next one jumps to the index of the previous log, which is why it is skipped In short, create copies of the list but check for the original, delete and add to the copy list and return the copies polecam rowniez zapoznac sie z funkcją list.contains()

Thomas

Poziom 41 , Bayreuth, Germany

14 kwietnia 2022, 15:11

A copy of the list is not necessary. You only need to keep track of the index. If nothing is deleted or added from the list, the index is increased by +1 as usual (just proceed to the next element). If you delete the element at the current index, then the index must not be increased to go to the next element. If the copy of the element at the current index is added at the same position, one must advance two indexes. That's all and that's exactly what Nicodem is doing. He just has some problems in his if - else if causes. Of course you could use a copy as well. But as said, it is not necessary and it is not the approach that has been chosen here. As I do not know what a pentas is I even can't tell if your first sentence is wrong or not. Till the word pentas it is wrong so maybe I missunderstand what you're telling here. You can remove elements while iterating over a list. You just can not use lista.remove() when using an iterator (). This includes the for-each loop since it implicitly uses an iterator.

ArrayList<String> lista = new ArrayList<String>();
Collections.addAll(lista, "a", "b", "c", "d");

for (int i = lista.size() - 1; i != -1; i--) // just so you can see an index
    lista.remove(i);

lista.forEach(System.out::println);

in the result should look similar to:

public static ArrayList<String> napraw(ArrayList<String> lista) {
        for (int i = 0; i < lista.size();) {
            if (s.contains("r") && s.contains("l")) {
                i++;
            } else if (s.contains("r")) {
                lista.remove(i);
            } else if (s.contains("l")) {
                lista.add(i, lista.get(i));
                i += 2;
            } else {
                i++;
            }
        }
        return lista;
    }
</co

Thomas

Poziom 41 , Bayreuth, Germany

14 kwietnia 2022, 12:18

Yes, a little hint ;) The conditional branching does not work as you may expect. If a and b are not equal to zero (line 44), then the code in the curly brackets after the if is executed (line 45 - nothing). Then the code continues at line 54 whilst you probably want to evaluate lines 46 and 49. Make this lines if conditionals and move them into the curly brackets in line 45.

if (a != 0 && b != 0) {
    if (a > 0) {
        lista.remove(i);
        i--;
    }
    if (b > 0) {
        lista.add(lista.get(i));
    }
}

or remove the outer conditional entirely as it is superfluous cause the lists size changes over time, you should not loop over a fixed length (c) but use list.size() inside the loops definition

for (int i = 0; i < lista.size(); i++) {

c-- has to be changed to i-- (as I already did above). From here start testing again... good luck

Guadalupe Gagnon

Poziom 37 , Tampa, United States

14 kwietnia 2022, 15:51

couple things: - The condition at line 44 is necessary because if both 'r' and 'l' are present (mean a && b are both not 0) then the word should remain unchanged in the list. This condition skips the next two if true. - at line 32 looping as long as 'i' is less than 'c' is also necessary because is you used list.size() then the loop would continue infinitely when items are added. This is because the code adds words to the end of the list and if they met the requirements to be added in the first place then when the loop revisits that item it will be added again, and again, and again. You 1/2 nailed the problem with changing line 48 to 'i--' instead of 'c--'. The thing is the code should also keep the 'c--' as when an item is removed you want to reduce the current index (which will increase by 1 at the start of the new loop cycle) AND decrease the effective end of the original list, which is what the variable 'c' is storing as value.

Thomas

Poziom 41 , Bayreuth, Germany

14 kwietnia 2022, 18:16

true with line 44. Somehow I skipped that condition. But you can loop over list.size() cause you insert at the current index (lista.add(i)) then you increment i by one right after that addition and another time in the loops definition... that way you skip the added element and hence no endless loop will occur. When removing an element you just do not increment at all (here you have to i--). In my above solution I do not increment in the loops definition but do everything when adding or removing elements. That way it seems to be clearer to me.

Guadalupe Gagnon

Poziom 37 , Tampa, United States

14 kwietnia 2022, 18:36

Ahh. I see. In your code that works because of the change to adding values at the current index. I missed that part of what you shared. In the original code, because values are added at the end of the list, that wouldn't work.

Thomas

Poziom 41 , Bayreuth, Germany

14 kwietnia 2022, 19:14

Yep, I just copy/ pasted the orig code to hint Nixodem into the right direction but leave some work to be done. As he had such a solution in mind I thought that shouldn't be that much of a problem to get the drift....but somehow I was tempted to reveal a little more.