How to take input in java without spaces?

Question

How can i take input for an integer array in a single line without spaces. Number can be only from 0-9 so every thing is single digit.

Input:-

Output:-

a[0]=1 a[1]=2 a[2]=3 a[3]=4 a[4]=5 a[5]=6

Please help me with this issue.

Are you just interested in reading single digits? If not, how do you plan to distinguish `123456` being `1,2,3,4,5,6` vs `12,34,56`? — azurefrog, Sep 05 '14 at 15:40
How would you know where one integer ends and the next one begins? This is only possible if you assume constant length for each int. — Eran, Sep 05 '14 at 15:41
@Reimeus I have no idea on what function to use i am used to taking inputs with spaces in between! — Adarsh, Sep 05 '14 at 15:41
@Audi : a simple nextInt() call on a Scanner can read a line of int separated by spaces, otherwise just use nextLine() and retrieve the digits by hand. — Dici, Sep 05 '14 at 15:42

score 6 · Answer 1 · edited May 23 '17 at 10:27

Assuming you are getting input as String you could do something like (streams and lambdas are available since Java 8)

String input = "123456";
int[] array = input.chars().map(i->i-'0').toArray();

System.out.println(Arrays.toString(array));

Output: [1, 2, 3, 4, 5, 6]

This code can be considered as shorter version of

String input = "123456";
int[] array = new int[input.length()];
int i = 0;
for (char c : input.toCharArray())
    array[i++] = c - '0';

Anyway this solution is based on idea that integer value or char represents its index in Unicode table. Since digits are ordered as 0,1,2,3,4,5,6,7,8,9 we are interested in differences between position of checked character and position of '0'. For example '3' is placed at index 51 while index of '0' is 48 so int value = '3'-'0' is translated as int value = 51 - 48 which creates integer with value 3.

Or if that is an option you could get char[] array instead of int[] which will contain all characters from string by simply using toCharArray

String input = "123456";
char[] array = input.toCharArray();

If you are also interested in result as String[] then since Java 8 you can simply use

String input = "123456";
String[] array = input.split("");

instead of cryptic (even for those with basic regex knowledge): split("(?!^)"), split("(?<!^)"), split("(?<=\\d)") or whatever regex you can think of.

Yours also works but i need a simpler solution so i prefer the previous one no offense. :) thanks:) — Adarsh, Sep 05 '14 at 15:59
@Audi What do you mean by simpler? IMHO `toCharArray` is most descriptive and simplest solutions. Also if you are interested in `String[]` result then as stated in my answer you can simply use `split("")` (this feature was added precisely to avoid regexes like `split("(?!^)")`). — Pshemo, Sep 05 '14 at 16:10

score 2 · Accepted Answer · answered Sep 05 '14 at 15:49

2

Take in in as a String and use split() -method like this:

"123".split("(?!^)")

will produce an array of strings

array ["1", "2", "3"]

answered Sep 05 '14 at 15:49

billerby

1,977
12
22

score 0 · Answer 3 · answered Sep 05 '14 at 15:48

If you assume your integers are all single-digit, you can do the following, but I don't see the interest of doing so :

 Scanner sc     = new Scanner(System.in);
 String  s      = sc.nextLine();
 int[]   values = new int[s.length()]; 
 for (int i=0 ; i<values.length ; i++) 
     values[i] = s.charAt(i) - '0';

score 0 · Answer 4 · answered Sep 05 '14 at 15:54

One way to do this is to realize that while the Scanner class parses things using a separator by default, there is a method called findInLine() which "Attempts to find the next occurrence of the specified pattern ignoring delimiters."

You can create a Pattern which matches a single digit, and then ask the Scanner to parse through sans-separator like so:

Scanner scan = new Scanner("123456");
Pattern digit = Pattern.compile("\\d");
String nextInt = scan.findInLine(digit);
while (nextInt != null) {
    System.out.println(nextInt);
    nextInt = scan.findInLine(digit);
}


output:
1
2
3
4
5
6

score 0 · Answer 5 · answered Sep 05 '14 at 17:01

0

try this

"123456".toCharArray()

answered Sep 05 '14 at 17:01

Genjuro

7,405
7
41
61

How to take input in java without spaces?

5 Answers5