使用Luhn算法检查信用卡有效性

Question

我尝试使用Luhn算法检查信用卡的验证,该算法的工作原理如下:

1. 从右到左加倍每秒.如果数字加倍会产生两位数字,请将两位数相加以得到一位数字.

   2*2 = 4

   2*2 = 4

   4*2 = 8

   1*2 = 2

   6*2 = 12(1 + 2 = 3)

   5*2 = 10(1 + 0 = 1)

   8*2 = 16(1 + 6 = 7)

   4*2 = 8

2. 现在添加步骤1中的所有单位数字.

   4 + 4 + 8 + 2 + 3 + 1 + 7 + 8 = 37

3. 在卡号中从右到左添加奇数位的所有数字.

   6 + 6 + 0 + 8 + 0 + 7 + 8 + 3 = 38

4. 对步骤2和步骤3的结果求和.

   37 + 38 = 75

5. 如果步骤4的结果可被10整除,则卡号有效; 否则,它无效.例如,号码4388576018402626无效,但号码4388576018410707有效.

简单地说,我的程序始终显示对我输入的所有内容都有效.即使它是有效数字,sumOfOddPlace和sumOfDoubleEvenPlace方法的结果也等于零.
任何帮助表示赞赏.

import java.util.Scanner;
public class CreditCardValidation {
      public static void main(String[] args) {
     Scanner in = new Scanner(System.in);
        int count = 0;
        long array[] = new long [16];
       do
       {
        count = 0;
       array = new long [16];
        System.out.print("Enter your Credit Card Number : ");
        long number = in.nextLong();
        for (int i = 0; number != 0; i++) {
        array[i] = number % 10;
        number = number / 10;
        count++;
        }
       }
        while(count < 13); 
        if ((array[count - 1] == 4) || (array[count - 1] == 5) || (array[count - 1] == 3 && array[count - 2] == 7)){
            if (isValid(array) == true) {
                System.out.println("\n The Credit Card Number is Valid. ");
        } else {
            System.out.println("\n The Credit Card Number is Invalid. ");
        }
        } else{
          System.out.println("\n The Credit Card Number is Invalid. ");
        }
    }

    public static boolean isValid(long[] array) {
        int total = sumOfDoubleEvenPlace(array) + sumOfOddPlace(array);        
        if ((total % 10 == 0)) {
         for (int i=0; i< array.length; i++){
            System.out.println(array[i]);}
            return true;
        } else {
          for (int i=0; i< array.length; i++){
            System.out.println(array[i]);}
            return false;
        }
    }

    public static int getDigit(int number) {
        if (number <= 9) {
            return number;
        } else {
            int firstDigit = number % 10;
            int secondDigit = (int) (number / 10);
            return firstDigit + secondDigit;
        }
    }

    public static int sumOfOddPlace(long[] array) {
        int result = 0;
        for (int i=0; i< array.length; i++)
        {
        while (array[i] > 0) {
            result += (int) (array[i] % 10);
            array[i] = array[i] / 100;
         }}
         System.out.println("\n The sum of odd place is " + result);
        return result;
    }

    public static int sumOfDoubleEvenPlace(long[] array) {
        int result = 0;
        long temp = 0;
        for (int i=0; i< array.length; i++){
        while (array[i] > 0) {
             temp = array[i] % 100;
             result += getDigit((int) (temp / 10) * 2);
            array[i] = array[i] / 100;
           }
        }
        System.out.println("\n The sum of double even place is " + result);
        return result;
    }
     }

Answer 1

您可以自由导入以下代码:

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);
    }
}

链接参考:https://github.com/jduke32/gnuc-credit-card-checker/blob/master/CCCheckerPro/src/com/gnuc/java/ccc/Luhn.java

Answer 2

谷歌和维基百科是你的朋友.我将使用int-array而不是long-array.在Wikipedia上发布以下java代码(以及Luhn算法的详细说明):

   public static boolean check(int[] digits) {
     int sum = 0;
     int length = digits.length;
     for (int i = 0; i < length; i++) {

       // get digits in reverse order
       int digit = digits[length - i - 1];

       // every 2nd number multiply with 2
       if (i % 2 == 1) {
           digit *= 2;
       }
       sum += digit > 9 ? digit - 9 : digit;
     }
     return sum % 10 == 0;
   }

您应该处理输入处理代码.我建议你研究以下解决方案:

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    boolean repeat;
    List<Integer> digits = new ArrayList<Integer>();

    do {
        repeat = false;
        System.out.print("Enter your Credit Card Number : ");
        String input = in.next();

        for (int i = 0; i < input.length(); i++) {
            char c = input.charAt(i);
            if (c < '0' || c > '9') {
                repeat = true;
                digits.clear();
                break;
            } else {
                digits.add(Integer.valueOf(c - '0'));
            }
        }
    } while (repeat);

    int[] array = new int[digits.size()];
    for (int i = 0; i < array.length; i++) {
        array[i] = Integer.valueOf(digits.get(i));
    }
    boolean valid = check(array);
    System.out.println("Valid: " + valid);
}

Answer 3

为简单起见，我将使用 5 位卡号。假设您的卡号是12345；如果我正确地读取了代码，您将存储在array各个数字中：

array[] = {1, 2, 3, 4, 5}

既然你已经有了数字，sumOfOddPlace你应该做类似的事情

public static int sumOfOddPlace(long[] array) {
    int result = 0;
    for (int i = 1; i < array.length; i += 2) {
        result += array[i];
    }
    return result;
}

并在sumOfDoubleEvenPlace：

public static int sumOfDoubleEvenPlace(long[] array) {
    int result = 0;
    for (int i = 0; i < array.length; i += 2) {
        result += getDigit(2 * array[i]);
    }
    return result;
}