i'm not getting it.

Question about the task

Under discussion

pls help me.

1. Create a list of words and populate it yourself.
2. The fix method should:
2.1. remove all words containing the letter "r" from the list
2.2. duplicate all words containing the letter "l".
2.3. if a word contains both "r" and "l", then leave it unchanged.
2.4. don't do anything to other words.

For example:
rose lyre love

Output:
lyre love love

Requirements:

The program must not read data from the keyboard.
The fix method should remove all words containing the letter "r" from the list. There is an exception: words containing both "r" and "l" should be left alone.
The fix method must duplicate words containing the letter "l" (add another instance of this word to the list). There is an exception: words containing both "l" and "r" should not be duplicated.
The fix method should not do anything with words that contain both "l" and "r".
The fix method should not do anything with words that don't contain either "l" or "r".

package com.codegym.task.task07.task0716; import java.util.ArrayList; /* R or L */ public class Solution { public static void main(String[] args) throws Exception { ArrayList<String> list = new ArrayList<String>(); list.add("rose"); // 0 list.add("love"); // 1 list.add("lyre"); // 2 list = fix(list); for (String s : list) { System.out.println(s); } } public static ArrayList<String> fix(ArrayList<String> list) { // write your code here //ArrayList<String> list1 = new ArrayList<String>(); for(int i=0; i<list.size(); i++){ String s = list.get(i); if(s.contains("r") && s.contains("l")) //System.out.println(""); continue; if(s.contains("r") && !s.contains("l")){ list.remove(i); i--; } if(!s.contains("r") && s.contains("l")){ //list1.add(list.get(i)); list.add(i++,list.get(i)); } } // list1.addAll(list) ; return list; } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

4 June 2020, 17:59

Line 42 is the problem. The arguments of the add statement are 2 separate statements: i++ and list.get(i). In this code the first word to be reached is rose, which will be removed correctly shifting down the remaining elements. This part works as intended. Now the list will have the words {"love", "lyre"} remaining and i will equal 0 again. Now the word "love" is checked. Because it has an 'l' but no 'r', it will reach line 42. This is what happens: - first index i (equal to 0) will be accepted as the first argument of the add() method to place the new word at. - next the ++ part will increase i to 1 << this specifically is the problem - next list.get(i) will get the word at index 1, which is "lyre" - "lyre" is added to list at index 0, resulting list is {"lyre", "love", "lyre"} - loop moves to i = 2 and checks "lyre" at index 2, contains both 'l' and 'r' - loop moves to i = 3 and breaks because 3 is equal to the size of the list. The output is now: lyre love lyre But lyre should not have been duplicated while love should have been

Tata

Level 16 , Hisar, Haryana, India

5 June 2020, 03:02

i did it like this but still not getting it and don't know how do it:

if(!s.contains("r")  && s.contains("l")){
//list1.add(list.get(i));

list.add(list.get(i));
                i++;

Tata

Level 16 , Hisar, Haryana, India

5 June 2020, 03:28

now i got it i did it like

list.add(i,list.get(i));
i++ ;

Thank you and namaste

Guadalupe Gagnon

Level 37 , Tampa, United States

5 June 2020, 04:55

=) I was going to say that you still wanted to put it in the spot that the old code was putting it at, just not increase the index number until after the full .add() method had completed. Glad you figured that one out.