How to read multiple integer values from one line in Java using BufferedReader object?

Question

I am using BufferedReader class to read inputs in my Java program. I want to read inputs from a user who can enter multiple integer data in single line with space. I want to read all these data in an integer array.

Input format- the user first enters how many numbers he/she want to enter

And then multiple integer values in the next single line-

INPUT:

5

2 456 43 21 12

Now, I read input using an object (br) of BufferedReader

int numberOfInputs = Integer.parseInt(br.readLine());

Next, I want to read next line inputs in an array

int a[] = new int[n];

But we cannot read using this technique

for(int i=0;i<n;i++)
{
   a[i]=Integer.parseInt(br.readLine()); //won't work
}

So, is there any solution to my problem or we can't just read multiple integers from one line using BufferedReader objects

Because using Scanner object we can read this type of input

for(int i=0;i<n;i++)
{
   a[i]=in.nextInt(); //will work..... 'in' is object of Scanner class
}

docs.oracle.com/javase/8/docs/api/java/lang/…

ajb
– ajb

2014-08-25 18:21:27 +00:00
Commented Aug 25, 2014 at 18:21 — ajb
– ajb, Commented Aug 25, 2014 at 18:21

Kumar · Accepted Answer · 2015-07-09 09:18:29Z

24

Try the next:

int a[] = new int[n];
String line = br.readLine(); // to read multiple integers line
String[] strs = line.trim().split("\\s+");
for (int i = 0; i < n; i++) {
    a[i] = Integer.parseInt(strs[i]);
}

edited Jul 9, 2015 at 9:18

Kumar

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answered Aug 25, 2014 at 18:21

Paul Vargas

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1 Comment

Paul Vargas Over a year ago

@LalitChoudhary You're welcome. Do not forget that if an answer solves your question satisfactorily, you can accept it.

Pramod · Accepted Answer · 2020-12-07 05:32:03Z

11

Late to the party but you can do this in one liner in Java 8 using streams.

InputStreamReader isr= new InputStreamReader();
BufferedReader br= new BufferedReader(isr);

int[] input = Arrays.stream(br.readLine().split("\\s+")).mapToInt(Integer::parseInt).toArray();

edited Dec 7, 2020 at 5:32

answered Sep 8, 2018 at 14:39

Pramod

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Comments

Kumar · Accepted Answer · 2015-07-09 08:52:11Z

5

If you want to read integers and you didn't know number of integers

String[] integersInString = br.readLine().split(" ");
int a[] = new int[integersInString.length];
for (int i = 0; i < integersInString.length; i++) {
    a[i] = Integer.parseInt(integersInString[i]);
}

answered Jul 9, 2015 at 8:52

Kumar

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Comments

Debu Shinobi · Accepted Answer · 2019-11-12 08:12:20Z

2

Integer.parseInt(br.readLine()) -> Reads a whole line and then converts it to Integers.

scanner.nextInt() -> Reads every single token one by one within a single line then tries to convert it to integer.

String[] in = br.readLine().trim().split("\\s+"); 
// above code reads the whole line, trims the extra spaces 
// before or after each element until one space left, 
// splits each token according to the space and store each token as an element of the string array.


for(int i = 0; i < n; i++)
    arr[i] = Integer.parseInt(in[i]);

// Converts each element in the string array as an integer and stores it in an integer array.

answered Nov 12, 2019 at 8:12

Debu Shinobi

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Comments

Saeed Zhiany · Accepted Answer · 2018-11-09 07:09:17Z

1

import java.io.*;
public class HelloWorld{

     public static void main(String []args){
        int i;
        System.out.println("enter the array element");
        InputStreamReader isr= new InputStreamReader();
        BufferedReader ib= new BufferedReader(isr);
        int a[]=new int [5];
        for(i=0;i<5;i++)
        {
            a[i]= Integer.parseInt(ib.readLine(a[i]));

        }
        for(i=0;i<5;i++)
        {
            System.out.println(a[i]);
        }

     }
}

edited Nov 9, 2018 at 7:09

Saeed Zhiany

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answered Nov 9, 2018 at 7:00

alok

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1 Comment

cezar Over a year ago

Please add some explanation. The answer should be useful to a broader public, not just to the OP.

ChillyWe · Accepted Answer · 2019-07-16 11:15:22Z

1

I use this code for List:

List<Integer> numbers = Stream.of(reader.readLine().split("\\s+")).map(Integer::valueOf).collect(Collectors.toList());

And it is almost the same for array:

int[] numbersArray = Arrays.stream(reader.readLine().split("\\s+")).mapToInt(Integer::valueOf).toArray();

edited Jul 16, 2019 at 11:15

answered Jul 16, 2019 at 11:09

ChillyWe

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Comments

VaibhavDomadia · Accepted Answer · 2018-10-25 14:49:09Z

0

You can use StringTokenizer class of java.util package. The StringTokenizer class allows an application to break a string into tokens. You can use this tokens using nextToken() method of StringTokenizer class.

You can use following constructor of StringTokenizer:

StringTokenizer(String str, String delimiter);

you can use space(" ") as delemeter.

int a[] = new int[N];
StringTokenizer st = new StringTokenizer(br.readLine() , " ");
for(int i=0 ; i<N ; i++) {
    a[i] = Integer.parseInt(st.nextToken());
}

answered Oct 25, 2018 at 14:49

VaibhavDomadia

1

Comments

JP. Aulet · Accepted Answer · 2020-12-18 19:30:35Z

0

Well as you mentioned there are two lines- First line takes number of integers and second takes that many number

INPUT: 5 2 456 43 21 12

So to address this and covert it into array -

String[] strs = inputData.trim().split("\\s+");  //String with all inputs 5 2 456 43 21 12

int n= Integer.parseInt(strs[0]);    //As you said first string contains the length

int a[] = new int[n];               //Initializing array of that length

for (int i = 0; i < n; i++)         
{

a[i] = Integer.parseInt(strs[i+1]);     // a[0] will be equal to a[1] and so on....

}

edited Dec 18, 2020 at 19:30

JP. Aulet

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answered Dec 18, 2020 at 18:48

Ragner16

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Comments

Abhay kolhe · Accepted Answer · 2020-12-21 17:51:42Z

0

i/p:34 54

 BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                                                                        
 StringTokenizer st = new StringTokenizer(br.readLine()," ");
 for(int i=0;i<st.countTokens();i++){
   a=Integer.parseInt(st.nextToken());
   b=Integer.parseInt(st.nextToken());
 }

//For fast input output for one line

answered Dec 21, 2020 at 17:51

Abhay kolhe

1

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