首先，这是一个关于信用卡号验证的学校项目。 基本上我试图将long int（CC编号）转换为向量，这样我就可以根据需要操作每个数字。我是通过使用％10方式来实现的。因为这将导致我的向量具有从右到左（向后）的数字，所以我从ccNum.size（） - 1开始以增量设置向量并向后工作，以便向量中的值具有相同的顺序作为用户输入。 问题是处理的第一个值（所以输入的最后9位左右）被错误地放入向量中。我已经尝试重构循环，复制到数组然后将其反转到向量中，以及我脑海中无法跟踪的各种其他事情。我试图尽可能地评论我的代码，以概述正在发生的事情，这里是：
public static void main(String[] args) {
long creditCardNumber = 0;
Vector<Integer> ccNum = new Vector<Integer>(13,3);
Vector<Integer> oddNum = new Vector<Integer>(6,1);
Vector<Integer> evenNum = new Vector<Integer>(6,1);
creditCardNumber = getInput(creditCardNumber);
int size = getSize(creditCardNumber);
//Pre-populate the vector so that I can add values starting from behind
//this makes it so that the credit card number isn't backwards. I also
//did it so that it matches the size of the input to prevent false values
//Yes, I verified the getSize method works.
for (int k = 0; k < size; k++) {
ccNum.add(k);
}
//I made a copy of the variable so I don't screw it up in the loop
long ccNumber = creditCardNumber;
//THIS IS WHERE THE PROBLEM IS.
for (int j = 0; ccNumber > 0; j++){
//on the first iteration, set the final index of ccNum array to
//final value of input using %10
if (j == 0){
ccNum.set(ccNum.size() - 1, (int) ccNumber % 10);
ccNumber /= 10;
//This just prints the value of ccNumber afterwards so that I can
//keep track of whats going on and make sure everything is "working"
System.out.println(ccNumber);
} else {
//Here I am continuously setting the values going backwards
//I end up with the same amount of values as the input but some
//are wrong
ccNum.set(ccNum.size() - (j+1), (int) ccNumber % 10);
ccNumber /= 10;
System.out.println(ccNumber);
}
}
//Here is where I print out all the values in the ccNum vector.
for (int l = 0; l < size; l++) {
System.out.print(ccNum.get(l));
}
System.out.println("");
for (int i = 0; i < ccNum.size() - 1; i ++) {
if (i % 2 == 0) {
evenNum.add(getDigit(ccNum.get(i)));
} else {
oddNum.add(ccNum.get(i));
}
}
int prefix = getPrefix(ccNum);
int sumEven = sumOfDoubleEvenPlace(evenNum);
int sumOdd = sumOfOddPlace(oddNum);
if (isValid(ccNum, sumEven, sumOdd, size, prefix)) {
System.out.printf("%d: is valid", creditCardNumber);
} else {
System.out.printf("%d: is invalid", creditCardNumber);
}
}
private static int sumOfOddPlace(Vector<Integer> oddNum) {
int sum = 0;
for (int i = 0; i < oddNum.size() - 1; i++) {
sum += oddNum.get(i);
}
return sum;
}
private static int sumOfDoubleEvenPlace(Vector<Integer> evenNum) {
int sum = 0;
for (int i = 0; i < evenNum.size() - 1; i++) {
sum += evenNum.get(i);
}
return sum;
}
private static int getDigit(int x) {
int y = x*2;
if (y < 10) {
return y;
} else {
int sum = 0;
while (y > 0) {
sum += y % 10;
y /= 10;
}
return sum;
}
}
private static int getPrefix(Vector<Integer> ccNum) {
if (ccNum.get(0) == 4) {
return 4;
} else if (ccNum.get(0) == 5) {
return 5;
} else if (ccNum.get(0) == 6) {
return 6;
} else if (ccNum.get(0) == 3 && ccNum.get(1) == 7) {
return 37;
} else {
return 0;
}
}
private static int getSize(long creditCardNumber) {
int size = (int)(Math.log10(creditCardNumber)+1);
return size;
}
private static long getInput(long creditCardNumber){
Scanner input = new Scanner(System.in);
System.out.print("Enter a credit card number: ");
creditCardNumber = input.nextLong();
input.close();
return creditCardNumber;
}
private static boolean isValid(Vector<Integer> ccNum, int sumEven, int sumOdd, int size, int prefix) {
if (size < 13 || size > 16) {
return false;
} else if (prefixMatched(prefix) == false) {
return false;
} else if ((sumEven + sumOdd) % 10 != 0) {
return false;
} else {
return true;
}
}
private static boolean prefixMatched(int prefix) {
if (prefix == 4 || prefix == 5 || prefix == 6 || prefix == 37) {
return true;
} else {
return false;
}
}
这是输出：
Enter a credit card number: 4388576018410707
438857601841070
43885760184107
4388576018410
438857601841
43885760184
4388576018
438857601
43885760
4388576
438857
43885
4388
438
43
4
0
438857601249-2-16-3
4388576018410707: is invalid
正如您所看到的，数字的前半部分是正确的（后半部分是循环的）。如果有人有一个很棒的解决方案。
对于可能会遇到困难的其他人的未来参考 - 这就是我最终做的事情，看起来更干净，感谢大家的帮助。
public class CreditCardNumberValidation {
public static void main(String[] args) {
long creditCardNumber = 0;
//Using ArrayLists because it makes it easier to manipulate data later
ArrayList<Integer> ccNum = new ArrayList<Integer>();
ArrayList<Integer> oddNum = new ArrayList<Integer>();
ArrayList<Integer> evenNum = new ArrayList<Integer>();
//Getting input in different method to clean up the code
creditCardNumber = getInput(creditCardNumber);
int size = getSize(creditCardNumber);
//Split creditCardNumber into separate integers and store in ArrayList
long ccNumber = creditCardNumber;
for (int j = 0; ccNumber > 0; j++){
ccNum.add((int) (ccNumber % 10));
ccNumber /= 10;
}
//Reverse the collection so that the numbers are in order
Collections.reverse(ccNum);
//Using the main List, even and odd numbers are sorted
for (int i = 0; i < ccNum.size(); i ++) {
if (i % 2 == 0) {
evenNum.add(getDigit(ccNum.get(i)));
} else {
oddNum.add(ccNum.get(i));
}
}
int prefix = getPrefix(ccNum);
int sumEven = sumOfDoubleEvenPlace(evenNum);
int sumOdd = sumOfOddPlace(oddNum);
if (isValid(ccNum, sumEven, sumOdd, size, prefix)) {
System.out.printf("%d is valid", creditCardNumber);
} else {
System.out.printf("%d is invalid", creditCardNumber);
}
}
private static int sumOfOddPlace(ArrayList<Integer> oddNum) {
int sum = 0;
for (int i = 0; i < oddNum.size(); i++) {
sum += oddNum.get(i);
}
return sum;
}
private static int sumOfDoubleEvenPlace(ArrayList<Integer> evenNum) {
int sum = 0;
for (int i = 0; i < evenNum.size(); i++) {
sum += evenNum.get(i);
}
return sum;
}
private static int getDigit(int x) {
int y = x*2;
if (y < 10) {
return y;
} else {
int sum = 0;
while (y > 0) {
sum += y % 10;
y /= 10;
}
return sum;
}
}
private static int getPrefix(ArrayList<Integer> ccNum) {
if (ccNum.get(0) == 4) {
return 4;
} else if (ccNum.get(0) == 5) {
return 5;
} else if (ccNum.get(0) == 6) {
return 6;
} else if (ccNum.get(0) == 3 && ccNum.get(1) == 7) {
return 37;
} else {
return 0;
}
}
private static int getSize(long creditCardNumber) {
//Easy way of getting the size of an integer without modifying it
return (int)(Math.log10(creditCardNumber)+1);
}
private static long getInput(long creditCardNumber){
Scanner input = new Scanner(System.in);
System.out.print("Enter a credit card number: ");
creditCardNumber = input.nextLong();
input.close();
return creditCardNumber;
}
private static boolean isValid(ArrayList<Integer> ccNum, int sumEven, int sumOdd, int size, int prefix) {
// Check size, prefix, and sum, if all tests pass return true
if (size < 13 || size > 16) {
return false;
} else if (prefixMatched(prefix) == false) {
return false;
} else if ((sumEven + sumOdd) % 10 != 0) {
return false;
} else {
return true;
}
}
private static boolean prefixMatched(int prefix) {
if (prefix == 4 || prefix == 5 || prefix == 6 || prefix == 37) {
return true;
} else {
return false;
}
}
}
答案 0 :(得分：0)
尝试替换
(int) ccNumber % 10
通过
(int) (ccNumber % 10)
如果先转换为int然后再执行模10，那么当int值溢出时它将无效。
答案 1 :(得分：-1)
我建议你在我的列表中使用ArrayList。我的函数返回Long中的整数的简单arraylist。
public List<Integer>getCardNumbers(Long longCardNumber){
String cardNumber = longCardNumber.toString();
List<Integer> intCardNumberList = new ArrayList<Integer>();
for(int i=0;i<cardNumber.length();i++){
intCardNumberList.add(Integer.parseInt(String.valueOf(cardNumber.charAt(i))));
}
return intCardNumberList;
}