Luhn algorithm (mod10) in Java

luhn.java

/**
 * Luhn Class is an implementation of the Luhn algorithm that checks validity of a credit card number.
 * 
 * @author <a href="http://www.chriswareham.demon.co.uk/software/Luhn.java">Chris Wareham</a>
 * @version Checks whether a string of digits is a valid credit card number according to the Luhn algorithm. 1. Starting with the second to last digit and
 *           moving left, double the value of all the alternating digits. For any digits that thus become 10 or more, add their digits together. For example,
 *           1111 becomes 2121, while 8763 becomes 7733 (from (1+6)7(1+2)3). 2. Add all these digits together. For example, 1111 becomes 2121, then 2+1+2+1 is
 *           6; while 8763 becomes 7733, then 7+7+3+3 is 20. 3. If the total ends in 0 (put another way, if the total modulus 10 is 0), then the number is valid
 *           according to the Luhn formula, else it is not valid. So, 1111 is not valid (as shown above, it comes out to 6), while 8763 is valid (as shown
 *           above, it comes out to 20).
 * @param ccNumber
 *            the credit card number to validate.
 * @return <b>true</b> if the number is valid, <b>false</b> otherwise.
 */
public class Luhn
{
        public static boolean Check(String ccNumber)
        {
                int sum = 0;
                boolean alternate = false;
                for (int i = ccNumber.length() - 1; i >= 0; i--)
                {
                        int n = Integer.parseInt(ccNumber.substring(i, i + 1));
                        if (alternate)
                        {
                                n *= 2;
                                if (n > 9)
                                {
                                        n = (n % 10) + 1;
                                }
                        }
                        sum += n;
                        alternate = !alternate;
                }
                return (sum % 10 == 0);
        }
}

n = (n % 10) + 1;

could be

n -= 9;

since n is less than 10 before being multiplied by 2

Hi Chris. Very nice and efficient algorithm you made. The most elegant approach to Luhn I have seen yet.

Hi, is there a library for this code that i can use in my project ?

@shibha: https://commons.apache.org/proper/commons-validator/apidocs/org/apache/commons/validator/routines/checkdigit/LuhnCheckDigit.html

n = (n % 10) + 1;

could be

n -= 9;

since n is less than 10 before being multiplied by 2

n = 1
n *= 2 => 2
n -=9 => -7 incorrect

n = (n % 10) + 1;

could be

n -= 9;

since n is less than 10 before being multiplied by 2

n = 1
n *= 2 => 2
n -=9 => -7 incorrect

hopefully, there is a precondition: if (n > 9)

;)