How to get the separate digits of an int number?

Question

I have numbers like 1100, 1002, 1022 etc. I would like to have the individual digits, for example for the first number 1100 I want to have 1, 1, 0, 0.

How can I get it in Java?

Answer 1

Convert it to String and use String#toCharArray() or String#split().

String number = String.valueOf(someInt);

char[] digits1 = number.toCharArray();
// or:
String[] digits2 = number.split("(?<=.)");

Answer 2

To do this, you will use the % (mod) operator.

int number; // = and int

while (number > 0) {
    print( number % 10);
    number = number / 10;
}

The mod operator will give you the remainder of doing int division on a number.

So,

10012 % 10 = 2

Because:

10012 / 10 = 1001, remainder 2

Note: As Paul noted, this will give you the numbers in reverse order. You will need to push them onto a stack and pop them off in reverse order.

Code to print the numbers in the correct order:

int number; // = and int
LinkedList<Integer> stack = new LinkedList<Integer>();
while (number > 0) {
    stack.push( number % 10 );
    number = number / 10;
}

while (!stack.isEmpty()) {
    print(stack.pop());
}

Answer 3

I'm not familiar with Java's internals, but couldn't you simply modulus the number by 10 repeatedly, and store the result in a variable?

Answer 4

Integer.toString(1100) gives you the integer as a string. Integer.toString(1100).getBytes() to get an array of bytes of the individual digits.

Edit:

You can convert the character digits into numeric digits, thus:

  String string = Integer.toString(1234);
  int[] digits = new int[string.length()];

  for(int i = 0; i<string.length(); ++i){
    digits[i] = Integer.parseInt(string.substring(i, i+1));
  }
  System.out.println("digits:" + Arrays.toString(digits));

Answer 5

Basically I agree with BalusC. But note that this gives characters, which may or may not be what you want. That is, you'll get the character '1' rather than 1. But this is easily converted by subtracting '0'. For example:

char[] charDigits=Integer.toString(n).toCharArray();
int[] digits=new int[charDigits.length];
for (int p=0;p<charDigits.length;++p)
  digits[p]=charDigits[p]-'0';
return digits;

Okay, this won't work if you have some Unicode digits that don't fall in the range of '0' to '9', but I'm not sure if there are any such, and in any case I don't think you're going to get them out of Integer.toString.

Answer 6

How about this?

public static void printDigits(int num) {
    if(num / 10 > 0) {
        printDigits(num / 10);
    }
    System.out.printf("%d ", num % 10);
}

or instead of printing to the console, we can collect it in an array of integers and then print the array:

public static void main(String[] args) {
    Integer[] digits = getDigits(12345);
    System.out.println(Arrays.toString(digits));
}

public static Integer[] getDigits(int num) {
    List<Integer> digits = new ArrayList<Integer>();
    collectDigits(num, digits);
    return digits.toArray(new Integer[]{});
}

private static void collectDigits(int num, List<Integer> digits) {
    if(num / 10 > 0) {
        collectDigits(num / 10, digits);
    }
    digits.add(num % 10);
}