CodeGym/Help with Java Tasks/Too much too many

Level 18

Kriva Palanka

03.05.2019
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14comments

Too much too many

Question about the task

Archived

When I print out the values if there is new line "println" verification says that there are too many, if I use "print" says there are too few. DFQ?

1. Read a file name from the console.
2. Read a set of numbers from the file.
3. Display only the even numbers, sorted in ascending order.

Example input:
5 8 -2 11 3 -5 2 10

Example output:
-2 2 8 10

Requirements:

The program must read data from the console.
The program must create a FileInputStream using the line read from the console.
The program should display data on the screen.
The program should display all of the even numbers read from the file, sorted in ascending order.
The program must close the input stream used to read the file (FileInputStream).

package com.codegym.task.task13.task1326; /* Sorting even numbers from a file */ import java.io.BufferedReader; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.Collections; public class Solution { public static void main(String[] args) { try { BufferedReader uin = new BufferedReader(new InputStreamReader(System.in)); String s = uin.readLine(); FileInputStream fis = new FileInputStream(s); ArrayList<Integer> list = new ArrayList<>(); while (fis.available() > 0) { int x = fis.read(); if (x % 2 == 0) list.add(x); } uin.close(); fis.close(); Collections.sort(list); for (int i = 0; i < list.size(); i++) { System.out.println(list.get(i)); } } catch (IOException e) { } } }

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Serghei

Level 26 , Moldova, Republic of

6 April 2020, 20:26

This code passed:

ArrayList<Integer> evenNumbers = new ArrayList<>();

 FileInputStream inputStream = new FileInputStream(new BufferedReader(new InputStreamReader(System.in)).readLine());
        BufferedReader input = new BufferedReader(new InputStreamReader(inputStream));

        String x;
        while ((x= input.readLine()) != null) {
            int n = Integer.parseInt(x);
            if (n % 2 == 0) {
                evenNumbers.add(n);
            }
        }

        input.close();
        inputStream.close();

        Collections.sort(evenNumbers);

        for (Integer n: evenNumbers) {
                System.out.println(n);
        }

Hưởng Nguyễn Công

Level 27 , Hanoi, Viet Nam

21 September 2019, 11:14

Hi, it tells us to use FileInputStream, and there was a trick here. We could also use Scanner to wrap FileInputStream, and every things else should be easy:

Scanner scanner = new Scanner(fileInputStream);
            while (scanner.hasNextInt()){
                int a = scanner.nextInt();
                //your code here
            }

Henk

Level 19 , Pretoria, South-Africa

16 May 2019, 13:49

But it tells us to use FileInputStream.... ?

Darko Jakimovski

Level 18 , Kriva Palanka, Macedonia, The Former Yugoslav Republic of

17 May 2019, 09:19

And so I do

Henk

Level 19 , Pretoria, South-Africa

17 May 2019, 10:28

no I am talking about the code of Guadalupe (our constant saviour in this course, LOL!) He's saying to use BR or Scanner, instead of FileInputStream. I tried EVERYTHING, cannot get mine to work, sheesh.

Darko Jakimovski

Level 18 , Kriva Palanka, Macedonia, The Former Yugoslav Republic of

17 May 2019, 10:58

Yeah but you can wrap the FileInputStream in the buffer initialization. Guadalupe owns this! I hope he got a job by now, he is the best among us wanderers

Henk

Level 19 , Pretoria, South-Africa

17 May 2019, 11:34

Yes, he does! I also see you have a nice grasp on things sometimes when I check your comments/code. Good luck! I have been a programmer and have a job for the last almost 24 years, but in Cobol and C. So the OO part is quite complex for me. Also, I hate that there's so much confusion with regard to how many different ways you can do things, and how different every one in Google's replies to questions are. It's a big wide complex world, lol. But we'll get there!

Darko Jakimovski

Level 18 , Kriva Palanka, Macedonia, The Former Yugoslav Republic of

17 May 2019, 11:53

I'm sure we will. "Thinking In Java" will help you get a firmer grasp on OO, highly recommended by everyone. javaranch is good place to discuss java as well. GL and HF Henk.

Guadalupe Gagnon

Level 37 , Tampa, United States

3 May 2019, 16:34

You are reading in individual bytes, so if the input is -15, what is read in will be: 45 // ascii for '-' 49 // ascii for '1' 53 // ascii for '5' You either need to code in to continue reading until a new line character is reached, or use a BufferedReader/Scanner. I suggest the BR/Scanner way as it is easier to understand and does exactly what you need. See changes below:

BufferedReader fis = new BufferedReader(new FileReader(s)); // change to BR instead of a FIS

ArrayList<Integer> list = new ArrayList<>();
while (fis.available() > 0) {
      int x = Integer.parseInt(fis.readLine()); // parse to int
      if (x % 2 == 0)
         list.add(x);
}

Guadalupe Gagnon

Level 37 , Tampa, United States

3 May 2019, 16:39

You could also read in everything at once and convert it to a string, then use that string to solve. The code would be:

FileInputStream fis = new FileInputStream(s);
ArrayList<Integer> list = new ArrayList<>();
byte[] temp = new byte[fis.available()]; // creates a byte array in exactly the size you need
fis.read(temp); // reads in all the data at once and stores it into the byte array
String result = new String(temp); // converts the byte array to string format

The file would now be stored in 'result' in full exactly as it was. You would just have to modify the bottom half of your code to split that String into the numbers and then sort them.

Guadalupe Gagnon

Level 37 , Tampa, United States

3 May 2019, 16:46

Me, personally, I prefer to use InputStreams and read everything at once. Every individual read takes a set amount of time to 1) set up, 2) read, and 3) return the data. Doing it only one time saves time on all three of those. When you end up reading very large files, the time saved by doing it this way is quite shocking.

Darko Jakimovski

Level 18 , Kriva Palanka, Macedonia, The Former Yugoslav Republic of

13 May 2019, 08:16

Thanks again man. You're the best. I tried everything doesn't verify. I give up for now.

Shndlr B

Level 17 , France

15 May 2019, 17:52

Try using it like this

new BufferedReader(new InputStreamReader(new FileInputStream(s)))

instead of using the raw FileInputStream. Then use this BufferedReader the same way you used System.in. I hope this isn't overkill.

Ahmed

Level 23 , Amsterdam, Netherlands

24 May 2019, 16:21

@Guadalupe, thanks. Reading everything at once did the trick for me. I was snuggling with this for almost two hours. Thanks again!