我的递归代码在Java中计算数字根有什么问题？ 我该如何解决？

Question

This is the first time I have tried recursion, so sorry for the stupidity. I am trying to calculate the digital root of a number by Java. I checked that we could calculate it by dividing by 9, but I still want to use recursion. Could you tell me what's wrong with my recursion code in Java? How can I fix it? Could you provide sample code for me?

public static void main(String[] args) {
    Scanner console = new Scanner(System.in);
    int inputnumber = inputnumber(console); 
    int sumofdigit = sumofdigit(inputnumber);
    int digitalroot = digitalroot(inputnumber);
    System.out.println("That number is :" + digitalroot);
}

//input console
public static int inputnumber(Scanner console){
    System.out.println("Please input: ");
    int num = console.nextInt();
    return num;}

public static int digitalroot(int inputnumber ) {
    if(inputnumber<10){
        return inputnumber;
    } else {
        return digitalroot(sumofdigit(inputnumber));
    }
}

// calculate sum of digits
public static int sumofdigit(int inputnumber){
    return sumofdigit(inputnumber/10) + inputnumber%10; 
}

Answer 1

Your recursion never ends. Try this with the added if statement:

public static int sumofdigit(int inputnumber) {
    if (inputnumber<10)
        return inputnumber;
    return sumofdigit(inputnumber/10) + inputnumber%10; 
}

You have a simlar snippet in digitalroot , but you primarily need it in sumofdigit .

Answer 2

I'd say it is slightly easier to just stop recursion once you hit zero (one less magic number, and it's more often the case that you can stop recursion at zero).

public static int sumofdigit(int inputnumber) {
    if (inputnumber == 0)
        return 0;
    return sumofdigit(inputnumber / 10) + inputnumber % 10;
}

Not that it matters much, but it at least tries to handle summing negative numbers (albeit returning a negated sum).

Answer 3

当您的inputnumber小于10时，sumofdigit方法应返回一个值以结束递归

Answer 4

Your problem is here:

public static int sumofdigit(int inputnumber){
        return sumofdigit(inputnumber/10) + inputnumber%10; 
}

To fix this you should do:

public static int sumofdigit(int inputnumber) {
    if (inputnumber<10)
        return inputnumber;
    return sumofdigit(inputnumber/10) + inputnumber%10; 
}

Here is some explanation on the problem and the solution:

As you might know, any recursion can be turned into a while loop. Your problematic recursion can be turned into

public static int sumofdigit(int inputnumber) {
    while(true){
      inputNumber = (inputNumber / 10) + (inputnumber%10); 
    }
    return inputNumber;
}

As you can see, this method will never return and will loop undefinitively. A quick fix would be to return when inputnumber<10

public static int sumofdigit(int inputnumber) {
    while(inputnumber>=10){
      inputNumber = (inputNumber / 10) + (inputnumber%10); 
    }
    return inputNumber;
}

Which turned back into recursion gives:

public static int sumofdigit(int inputnumber) {
        if (inputnumber<10)
            return inputnumber;
        return sumofdigit(inputnumber/10) + inputnumber%10; 
}

Answer 5

I wouldn't do this recursively either:

public static int sumofdigit(int x) {
    int sumofdigit = 0;
    while(x != 0) {
        sumofdigit += x % 10;
        x /= 10;
    }
    return sumofdigit;
}

public static int digitalRoot(int x) {
    while(x > 9) {
        x = sumofdigit(x);
    }
    return x;
}