Getting a wrong output using arraylists

Question

The challenge is to find a number whose individual digits multiplied by consecutively increasing power and added up, equal the initial number.

Eg: take 89, split it into 8 and 9, then 8^1 + 9^2 = 89

static List<Integer> sumDigPow(int a, int b) { 
        List<Integer> eureka = new ArrayList<Integer>(0);
        List<String> digits = new ArrayList<String>();
        String num;
        int sum = 0, multi;

    for (int i=a; i<=b; i++) {
        num = String.valueOf(i);
        digits.add(num);

        for (int j=0; j<digits.size(); j++) {
                multi = (int)Math.pow(Integer.parseInt(digits.get(j)), j+1);
                sum += multi;
        }

        if (sum == i) eureka.add(i);

        sum = 0;
        digits.clear();
    }

    return eureka;
}

With an input of 1 and 100 (the range), the output should be [1, 2, 3, 4, 5, 6, 7, 8, 9, 89], but I'm getting all of the numbers [1, 2 ... 100].

I've started learning java fairly recently and can't seem to find the issue in the code. Any hints would be greatly appreciated.

Answer 1

You can use the following:

static List<Integer> sumDigPow(int a, int b) {
    List<Integer> eureka = new ArrayList<Integer>(0);
    String num;
    int sum = 0, multi;

    for (int i = a; i <= b; i++) {
        num = String.valueOf(i);
        for (int j = 0; j < num.length(); j++) {
            multi = (int) Math.pow(Character.getNumericValue(num.charAt(j)), j + 1);
            sum += multi;
        }

        if (sum == i) {
            eureka.add(i);
        }
        sum = 0;
    }
    return eureka;
}

Explanation:

1. You were not checking the second digit of the number.
2. Loop over each character of the String num.
3. There is no need of the digits arraylist, you can just use the numeric value of the char.

Answer 2

Use char[] to split the numbers into digits as an array of characters (you are just adding the whole number as a single string to the list, not its individual digits):

...
    char[] digits;
...
    digits = String.valueOf(i).toCharArray();

Then if you subtract '0' from each char digit you automatically get the actual int value of the digit without having to invoke the Integer.parseInt method on a String, or any other parsing method:

    (int)Math.pow(digits[j] - '0', j + 1);

The full code would look like this:

static List<Integer> sumDigPow(int a, int b) {
    List<Integer> eureka = new ArrayList<Integer>();
    int sum = 0;
    char[] digits;

    for (int i = a; i <= b; i++) {
        digits = String.valueOf(i).toCharArray();
        for (int j = 0; j < num.length(); j++) 
            sum += (int)Math.pow(digits[j] - '0', j + 1);

        if (sum == i) eureka.add(i);
        sum = 0;
    }
    return eureka;
}