4
\$\begingroup\$

Was fiddling with ; trying to write a program for getting all Armstrong Numbers between 1 and 10_000_000.

Following is my working solution:

public class ArmstrongNumbers {

    public static void main(String[] args) {
        IntStream.range(1, 10_000_000)
                .filter((n) -> {
                    int c = 0, temp = n;
                    while (temp > 0) {
                        c += Math.pow(temp % 10, Integer.toString(n).length());
                        temp /= 10;
                    }
                    return c == n;
                }).forEach(System.out::println);
    }

}

Output:

1
2
3
4
5
6
7
8
9
153
370
371
407
1634
8208
9474
54748
92727
93084
548834
1741725
4210818
9800817
9926315

Why I'm here:

  • Can this be made shorter?
  • Can any other Java 8 concept be used here?
  • Can the use of the while loop be avoided and a stream be used somehow (given that it is more elegant)?
  • Review overall correctness

Here is a link to a list of Armstrong Numbers for verifying the outputs.

Please note that there is a follow up question.

\$\endgroup\$
3
\$\begingroup\$

Integer.toString(n).length() can be assigned to its own variable.

If you wanted to stream the while loop, I suppose you could split a toString() version of the number via toCharArray() and then calculate the value for each digit individually, then summing the result via the sum function.

I don't know whether that'd qualify as "more elegant", though. You would get rid of c and temp as variables by doing that.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.