Armstrong Number Checker in Java

Question

I am still somewhat of a beginner to Java, but I need help with my code. I wanted to write an Armstrong Number checker.

An Armstrong number is one whose sum of digits raised to the power three equals the number itself. 371, for example, is an Armstrong number because 3^3 + 7^3 + 1^3 = 371.

If I understand this concept correctly, then my code should work fine, but I don't know where I made mistakes. I would appreciate if you could help correct my mistakes, but still kind of stick with my solution to the problem, unless my try is completely wrong or most of it needs to change.

Here is the code:

public class ArmstrongChecker {
    
     boolean confirm = false;
     Integer input;
     String converter;
     int indices;
     int result = 1;
    
     void ArmstrongCheck(Integer input) {
        this.input = input;
        converter = input.toString();
        char[] array = converter.toCharArray();
        indices = array.length;
        result = (int) Math.pow(array[0], indices);
        
        for (int i = 1; i < array.length; i++) {
            result = result + (int) Math.pow(array[i], indices);
        }
        if (result == input) {
            confirm = true;
            System.out.println(confirm);
        } else {
            System.out.println(confirm);
        }
    }
}

For my tries I used '153' as an input. Thank you for your help!

Answer 1

You aren't summing the digits, but the numeric values of the characters representing them. You can convert such a character to its numeric value by subtracting the character '0':

int result = 0;
for(int i = 0; i < array.length; i++) {
    result = result + (int) Math.pow(array[i] - '0', indices);
}

Having said that, it's arguably (probably?) more elegant to read the input as an actual int number and iterate its digits by taking the reminder of 10 on each iteration. The number of digits itself can be calculated using a base-10 log.

int temp = input;
int result = 0;
int indices = (int) Math.log10(input) + 1;
while (temp != 0) { 
    int digit = temp % 10;
    result += (int) Math.pow(digit, indices);
    temp /= 10;
}