Sum of squares of digits of an integer in Java

Question

I am trying to write a recursive implementation of a method that takes a non-negative argument, and returns the sum of the squares of its digits. For example, sumSquareDigits(10) should return 1 and sumSquareDigits(103) should return 10.

This is my code :

public static int sumSquareDigits(int n) {
    if (n < 10) return n^2;
    return (((n % 10)^2) + sumSquareDigits(n/10));
    }

For example for an given integer 625, it would be something like:

(625 % 10)^2 + (62,5 % 10)^2 + (6,25)^2 = 5^2 + 2^2 + (6,25)^2

which of course is wrong because the last term should be 6 and not 6,25. What I am looking for is a way to truncate 6,25 so that it becomes 6.

How do we do this? And I'd appreciate any hints to implement this function better (I'm new in Java).

Thanks!

Answer 1

In Java, ^ is not "to the power" of, it is the bitwise XOR operator. To perform powers in Java use Math.pow . Bear in mind that Math.pow returns a double so you will need to cast it to an int if you only want a whole number. Eg

if (n < 10) return (int)Math.pow(n, 2);

Of course, as msandiford pointed out, if you only need to calculate powers of 2, it is probably easier to just multiply a number by itself. Eg

if (n < 10) return n * n;

Answer 2

The ^ is for the bitwise XOR operator. You could use Math.pow , and cast the results to int since Math.pow returns a double :

public static int sumSquareDigits(int n) {
    if (n < 10) return (int) Math.pow(n, 2);
    return (int)((Math.pow((n % 10), 2)) + sumSquareDigits(n/10));
}

Or since it's only squared, just multiply the base by itself:

public static int sumSquareDigits(int n) {
    if (n < 10) return n * n;
    return ((n % 10) * (n % 10)) + sumSquareDigits(n/10);
}