almost stuck :D

Question about the task

Resolved

Hello, I seem to be stuck here. My code is almost correct, however it duplicates wrong string. Instead of outputing lyre love love it outputs love lyre lyre Any suggestions?

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 for (int i = 0; i < list.size(); ) { if (list.get(i).contains("r")&&!list.get(i).contains("l")) { list.remove(i);} else if (list.get(i).contains("l")&&!list.get(i).contains("r")) { list.add(++i, list.get(i));} else i++; } return list; } }

Comments (2)

Popular
New
Old

You must be signed in to leave a comment

Sinisa

Level 11 , Banja Luka, Bosnia and Herzegovina

3 March 2021, 09:39

solution

I succeeded eventually:

for (int i = 0; i < list.size(); ) {
    if ((list.get(i).contains("r")&&(list.get(i).contains("l")))||(!list.get(i).contains("r")&&!list.get(i).contains("l"))) {
        i++;}
        else if (list.get(i).contains("r")) {
        list.remove(i);
        }
        else if (list.get(i).contains("l")) {
        list.add(i, list.get(i));
        i+=2;}

Guadalupe Gagnon

Level 37 , Tampa, United States

2 March 2021, 21:40

2 problems. #1 when an item is deleted from a list all the items that occur further down the list have their index number reduced by 1. Lets say the loop cycle was 'i' equals 1, and you had a simple list: {zero, one, two, three} If you removed the item at index 1, the resulting list would be: {zero, two, three} Then the loop would move to the next cycle, so i would equal 2, and it would check the item at index 2, which is three. However there is a new item in index 1 that is essentially skipped: {zero, two, three} You need to account for this properly #2 ++ preceding a variable will increase it by 1 before an expression is evaluated; following the variable will increase it after the expression is evaluated. So, in both of these expressions if 'i' equaled 5: i++ - 4 << this would result in 1 because i is increased after the expression ++i - 4 << this would result in 2 because i is increased before the expression and in both cases i would equal 6. When no expression is being evaluated it really makes no difference if the ++ is before or after the variable. This is important because on line 28 i is being used to evaluate the value in the list, then at line 29 i is increased (without an expression) inside the put() method. This means it is grabbing and putting the next item in the list and not the item that was evaluated on line 28.