Java - How to reverse and split a number by digit into an array?

Question

I'm learning java coming from a ruby background and self-teaching.

I am doing a problem that wants a number reversed and listed into an array by digit.

Ex 1234 -> [4,3,2,1]

In ruby I did this easily -

def digitize(n)
  n.to_s.reverse.split("").map(&:to_i)
end

In java, my code is not working, because clearly, I am doing something wrong.

public class Kata {
  public static int[] digitize(long n) {

  String s = String.valueOf(n);
  String r = reverse(s, s.length()-1);
  String[] array = r.split("");
  Float[] floats = Arrays.stream(array).map(Float::valueOf).toArray(Float[]::new);
  }

  return floats;
}

Is there a simpler way to do this with built-in methods like ruby? I tried doing everything separately. First variable s , I wanted to convert the number to a string. Second variable r , reversing that string. Third variable array , splitting the variable r into digits. Floats at the end was to convert the array of strings to digits.

Answer 1

public class Kata {
  public static int[] digitize(long n) {
    String numStr = String.valueOf(n);
    int digits[] = new int[numStr.length()];
    int counter = 0;

    for(int i = numStr.length() - 1; i >= 0; i--) {
      digits[i] = Integer.parseInt(Character.toString(numStr.charAt(counter++)));
    }

    return digits;
  }
}

Answer 2

Your method returns Float[] instead of int[] . Try this:

int[] ints = Arrays.stream(array).mapToInt(Integer::parseInt).toArray();
return ints;

Answer 3

Use the JDK:

int[] reversedDigits = Arrays.stream(new StringBuilder().append(n).reverse().toString().split("(?<=.)")).mapToInt(Integer::parseInt).toArray();

Not as terse as ruby. A little embarrassing actually, but 1 line nevertheless.

Answer 4

You could collect a List and then reverse it with Collections.reverse(List) and then convert that to an int[] . Something like,

public static int[] digitize(long n) {
    String s = String.valueOf(n);
    List<Integer> al = s.chars().map(ch -> Character.digit(ch, 10))
            .boxed().collect(Collectors.toList());
    Collections.reverse(al);
    return al.stream().mapToInt(Integer::intValue).toArray();
}

or exploit the String.length() to build the array directly like

public static int[] digitize(long n) {
    String s = String.valueOf(n);
    int[] out = new int[s.length()];
    IntStream.range(0, s.length()).forEach( //
            i -> out[i] = Character.digit(s.charAt(s.length() - i - 1), 10));
    return out;
}

or combine that with a map for the even shorter

public static int[] digitize(long n) {
    String s = String.valueOf(n);
    return IntStream.range(0, s.length()) //
            .map(i -> Character.digit(s.charAt(s.length() - i - 1), 10)).toArray();
}

Answer 5

This works for me.

import java.lang.StringBuffer;

public class Kata {
  public static float[] digitize(long n) {

  int[] ints= new StringBuffer(n.toString()).reverse().toString().split("(?!^)").mapToInt(Integer::parseInt).toArray();;

  return ints;
}