[英]How to sum the digits of two integers?
如何在单独的方法中使用以下方法将两个数字的整数相加? 我正在尝试教自己如何使用重载方法,但这开始使我感到困惑。 谢谢!
public static void sumNUmber(){
System.out.println("Enter a number");
Scanner in = new Scanner(System.in);
int num = in.nextInt();
int sum = 0;
while (num > 0) {
sum = sum + num % 10;
num = num / 10;
}
System.out.println(sum);
public static void main(String[] args) throws ClassNotFoundException {
System.out.println("Enter a number");
Scanner in = new Scanner(System.in);
int num = in.nextInt();
System.out.println(sumDigits(num));
}
public static int sumDigits(int num) {
int sum = 0;
while (num > 0) {
sum += num % 10;
num = num / 10;
}
return sum;
}
输出
Enter a number
234
9
您可能希望将核心算法放入单个函数中,然后调用两次,每个数字一次。 例如,
// Core algorithm.
public static int sumDigits(int num) {
int sum = 0;
while (num > 0) {
sum += num % 10;
num = num / 10;
}
return sum;
}
public static void sumNumber() {
Scanner in = new Scanner(System.in);
System.out.println("Enter a number");
int numA = in.nextInt();
System.out.println("Enter another number");
int numB = in.nextInt();
int totalDigits = sumDigits(numA) + sumDigits(numB);
System.out.println(totalDigits);
}
//Merry Xmas :D
public int sumNumbers(int num) {
int returnval;
if (num < 10) {
return num;
} else if (num < 100) {
returnval = Math.floor(num/10) + (num%10);
} else if (num < 1000) {
returnval = Math.floor(num/100) + Math.floor(num/10) + (num%10);
} //repeat as needed
return 0;
}
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