Java - Recursion sum of number and how it work

Question

I am trying to write a recursive function that when I call with number 5 for example then the function will calculate the sum of all digits of five. 1 + 2 + 3 + 4 + 5 = 15

The current code always returns 0, how can the amount each time the n?

public class t {
public static void main(String[] args) {

    System.out.println(num(5));
}

public static int num(int n) {

    int sum = 0;
    sum += n;
    if (n == 0)
        return sum;

    return num(n - 1);

}

}

thank you.

Solution 1

Instead of setting the sum to 0 you can -
Do this:

public int sumUp(int n){

    if (n==1)
        return 1;
    else
       return sumUp(n-1)+n;
}

Solution 2

The problem is you set the sum always 0.

public static void main(String[] args) {
    System.out.println(num(5, 0));
}

public static int num(int n, int sum) {
    if (n == 0) {
        return sum;
    }

    sum += n;
    return num(n - 1, sum);

}

Solution 3

public static int withRecursion(List<Integer> list) {

    int size = list.size();
    int a=0;

    if(list.isEmpty() == true) {
        return 0;

    }else {
        a = a + list.get(0) + withRecursion(list.subList(1, size)); 
        return  a;      
    }   

}

Comments

[zapl]

@KevinWallis and that's not even all you can do. return n <= 1 ? n : num(n -1) + n; does it in 1 line.

[Tilak Madichetti]

he just seems to be .. learning java ; why do you wanna confuse him with that anyways, i'll add it to my answer..