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

Answer 1

Since I don't see a method on this question which uses Java 8, I'll throw this in. Assuming that you're starting with a String and want to get a List<Integer>, then you can stream the elements like so.

List<Integer> digits = digitsInString.chars()
        .map(Character::getNumericValue)
        .boxed()
        .collect(Collectors.toList());

This gets the characters in the String as a IntStream, maps those integer representations of characters to a numeric value, boxes them, and then collects them into a list.

Answer 2

in Java, this is how you can separate digits from numbers and store them in an Array.

public static void main(String[] args) {
        System.out.println("Digits Array:: "+Arrays.toString(getNumberArr(1100)));
}

private static Integer[] getNumberArr(int number) {
    //will get the total number of digits in the number
    int temp = number;
    int counter = 0;

    while (temp > 0) {
        temp /= 10;
        counter++;
    }
    //reset the temp
    temp = number;

    // make an array
    int modulo;     //modulo is equivalent to single digit of the number.
    Integer[] numberArr = new Integer[counter];
    for (int i = counter - 1; i >= 0; i--) {
        modulo = temp % 10;
        numberArr[i] = modulo;  
        temp /= 10;
    }

    return numberArr;
}

Output:

Digits Array:: [1, 1, 0, 0]

Answer 3

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 4

I wrote a program that demonstrates how to separate the digits of an integer using a more simple and understandable approach that does not involve arrays, recursions, and all that fancy schmancy. Here is my code:

int year = sc.nextInt(), temp = year, count = 0;

while (temp>0)
{
  count++;
  temp = temp / 10;
}

double num = Math.pow(10, count-1);
int i = (int)num;

for (;i>0;i/=10)
{
  System.out.println(year/i%10);
}

Suppose your input is the integer 123, the resulting output will be as follows:

1
2
3

Answer 5

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

int number; // = some 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 6

This uses the modulo 10 method to figure out each digit in a number greater than 0, then this will reverse the order of the array. This is assuming you are not using "0" as a starting digit.

This is modified to take in user input. This array is originally inserted backwards, so I had to use the Collections.reverse() call to put it back into the user's order.

    Scanner scanNumber = new Scanner(System.in);
    int userNum = scanNumber.nextInt(); // user's number

    // divides each digit into its own element within an array
    List<Integer> checkUserNum = new ArrayList<Integer>();
    while(userNum > 0) {
        checkUserNum.add(userNum % 10);
        userNum /= 10;
    }

    Collections.reverse(checkUserNum); // reverses the order of the array

    System.out.print(checkUserNum);