Java reading multiple ints from a single line

问题:

I am working on a program and I want to allow a user to enter multiple integers when prompted. I have tried to use a scanner but I found that it only stores the first integer entered by the user. For example:

Enter multiple integers: 1 3 5

The scanner will only get the first integer 1. Is it possible to get all 3 different integers from one line and be able to use them later? These integers are the positions of data in a linked list I need to manipulate based on the users input. I cannot post my source code, but I wanted to know if this is possible.

回答1:

I use it all the time on hackerearth

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    String  lines = br.readLine();    

    String[] strs = lines.trim().split("\\s+");

    for (int i = 0; i < strs.length; i++) {
    a[i] = Integer.parseInt(strs[i]);
    }

回答2:

Scanner has a method called hasNext():

    Scanner scanner = new Scanner(System.in);

    while(scanner.hasNext())
    {
        System.out.println(scanner.nextInt());
    }

回答3:

Try this

public static void main(String[] args) {
    Scanner in = new Scanner(System.in); 
    while (in.hasNext()) {
        if (in.hasNextInt())
            System.out.println(in.nextInt());
        else 
            in.next();
    }
}

By default, Scanner uses the delimiter pattern "\p{javaWhitespace}+" which matches at least one white space as delimiter. you don't have to do anything special.

If you want to match either whitespace(1 or more) or a comma, replace the Scanner invocation with this

Scanner in = new Scanner(System.in).useDelimiter("[,\\s+]");

回答4:

You want to take the numbers in as a String and then use String.split(" ") to get the 3 numbers.

String input = scanner.nextLine();    // get the entire line after the prompt 
String[] numbers = input.split(" "); // split by spaces

Each index of the array will hold a String representation of the numbers which can be made to be ints by Integer.parseInt()

回答5:

If you know how much integers you will get, then you can use nextInt() method

For example

Scanner sc = new Scanner(System.in);
int[] integers = new int[3];
for(int i = 0; i < 3; i++)
{
    integers[i] = sc.nextInt();
}

回答6:

Here is how you would use the Scanner to process as many integers as the user would like to input and put all values into an array. However, you should only use this if you do not know how many integers the user will input. If you do know, you should simply use Scanner.nextInt() the number of times you would like to get an integer.

import java.util.Scanner; // imports class so we can use Scanner object

public class Test
{
    public static void main( String[] args )
    {
        Scanner keyboard = new Scanner( System.in );
        System.out.print("Enter numbers: ");

        // This inputs the numbers and stores as one whole string value
        // (e.g. if user entered 1 2 3, input = "1 2 3").
        String input = keyboard.nextLine();

        // This splits up the string every at every space and stores these
        // values in an array called numbersStr. (e.g. if the input variable is 
        // "1 2 3", numbersStr would be {"1", "2", "3"} )
        String[] numbersStr = input.split(" ");

        // This makes an int[] array the same length as our string array
        // called numbers. This is how we will store each number as an integer 
        // instead of a string when we have the values.
        int[] numbers = new int[ numbersStr.length ];

        // Starts a for loop which iterates through the whole array of the
        // numbers as strings.
        for ( int i = 0; i < numbersStr.length; i++ )
        {
            // Turns every value in the numbersStr array into an integer 
            // and puts it into the numbers array.
            numbers[i] = Integer.parseInt( numbersStr[i] );
            // OPTIONAL: Prints out each value in the numbers array.
            System.out.print( numbers[i] + ", " );
        }
        System.out.println();
    }
}

回答7:

You're probably looking for String.split(String regex). Use " " for your regex. This will give you an array of strings that you can parse individually into ints.

回答8:

This works fine ....

int a = nextInt(); int b = nextInt(); int c = nextInt();

Or you can read them in a loop

回答9:

Better get the whole line as a string and then use StringTokenizer to get the numbers (using space as delimiter ) and then parse them as integers . This will work for n number of integers in a line .

    Scanner sc = new Scanner(System.in);
    List<Integer> l = new LinkedList<>(); // use linkedlist to save order of insertion
    StringTokenizer st = new StringTokenizer(sc.nextLine(), " "); // whitespace is the delimiter to create tokens
    while(st.hasMoreTokens())  // iterate until no more tokens
    {
        l.add(Integer.parseInt(st.nextToken()));  // parse each token to integer and add to linkedlist

    }

回答10:

Using this on many coding sites:

• CASE 1: WHEN NUMBER OF INTEGERS IN EACH LINE IS GIVEN

Suppose you are given 3 test cases with each line of 4 integer inputs separated by spaces 1 2 3 4, 5 6 7 8 , 1 1 2 2

        int t=3,i;
        int a[]=new int[4];

        Scanner scanner = new Scanner(System.in);

        while(t>0)  
        {
            for(i=0; i<4; i++){
                a[i]=scanner.nextInt();
                System.out.println(a[i]);
            }   

        //USE THIS ARRAY A[] OF 4 Separated Integers Values for solving your problem
            t--;
        }

• CASE 2: WHEN NUMBER OF INTEGERS in each line is NOT GIVEN

      Scanner scanner = new Scanner(System.in);
  
      String lines=scanner.nextLine();
  
      String[] strs = lines.trim().split("\\s+");
  

  Note that you need to trim() first: trim().split("\\s+") - otherwise, e.g. splitting a b c will emit two empty strings first

      int n=strs.length; //Calculating length gives number of integers
  
      int a[]=new int[n];
  
      for (int i=0; i<n; i++) 
      {
          a[i] = Integer.parseInt(strs[i]); //Converting String_Integer to Integer 
          System.out.println(a[i]);
      }
  

回答11:

Using BufferedReader -

StringTokenizer st = new StringTokenizer(buf.readLine());

while(st.hasMoreTokens())
{
  arr[i++] = Integer.parseInt(st.nextToken());
}

回答12:

i know it's old discuss :) i tested below code it's worked

`String day = "";
 day = sc.next();
 days[i] = Integer.parseInt(day);`

回答13:

When we want to take Integer as inputs
For just 3 inputs as in your case:

import java.util.Scanner;
Scanner scan = new Scanner(System.in);
int a,b,c;
a = scan.nextInt();
b = scan.nextInt();
c = scan.nextInt();

For more number of inputs we can use a loop:

import java.util.Scanner;
Scanner scan = new Scanner(System.in);
int a[] = new int[n]; //where n is the number of inputs
for(int i=0;i<n;i++){
    a[i] = scan.nextInt();    
}

回答14:

created this code specially for the Hacker earth exam

 {  



 Scanner values = new Scanner(System.in);  //initialize scanner
 int[] arr = new int[6]; //initialize array 
 for (int i = 0; i < arr.length; i++) {
        arr[i] = (values.hasNext() == true ? values.nextInt():null); // it will read 
 the next input value


/* user enter =  1 2 3 4 5
  arr[1]= 1
  arr[2]= 2
  and soo on */ 

     '}