using recursion to print symmetric integer sequenc

Question

I am just starting to learn java on my own by reading a book and completing the exercises. One of the exercises on the book is asking me to write a method that prints symmetric integer sequences using recursion. For example:

writeSequence(1); //prints 1 
writeSequence(2); //prints 1 1
writeSequence(3); //prints 2 1 2
writeSequence(4); //prints 2 1 1 2
writeSequence(5); //prints 3 2 1 2 3
writeSequence(6); //prints 3 2 1 1 2 3

So far, my solution uses two method implementations and doesn't look very elegant:

public class Client {

    public static void main(String[] args) {
            writeSequence(1);
            writeSequence(2);
            writeSequence(3);
            writeSequence(4);
            writeSequence(5);
            writeSequence(6);
    }

    public static void writeSequence(int num) {

        writeSequence( (int) Math.round(num/2.0), "desc", true);
        if (num % 2 == 0) {
            writeSequence( (int) Math.round(num/2.0), "asc", true);
        } else {
            writeSequence( (int) Math.round(num/2.0), "asc", false);
        }
    }

    public static void writeSequence(int num, String direction, Boolean show_one) {
        if (num < 1) {
            throw new IllegalArgumentException("less than 1");
        } else if (num == 1) {
            if (show_one) {
                System.out.print(1 + " ");
            }
        } else {
            if (direction.equals("desc")) {
                System.out.print(num + " ");
                writeSequence(num-1, direction, show_one);
            } else {
                writeSequence(num-1, direction, show_one);
                System.out.print(num + " ");
            }
        }
    }

}

My question is: how do I solve this problem using only ONE method that must take only ONE parameter? Right now, I am implementing two methods but I only want to implement one to solve the problem.

Answer 1

Here's a hint about how to write only one recursive method with one argument.

Print your number both before and after the recursive call in your recursive method.

Answer 2

I might be reading the same book as you were and I came upon this. I wasn't able to do it in one method, but I thought I'd post the code anyway to show a different way.

public static void writeSequence(int n) {

    int num = (int)Math.round(n / 2.0);
    String flag = "";
    if (n % 2 == 0) {
        flag = "even";
    } else {
        flag = "odd";
    }
    System.out.println(writeSequence(num, flag));
}

private static String writeSequence(int n, String flag) {
    if (n < 1) {
        throw new IllegalArgumentException("less than one");
    } else if (n == 1 && flag.equals("odd")) {
        return "1";
    } else if (n == 1 && flag.equals("even")) {
        return "1 1";
    } else { 
        return n + " " + writeSequence(n - 1, flag) + " " + n;
    }
}

Answer 3

I know it was an old post, but I've just literally went through this problem on praticeit and here is my solution

public void writeSequence(int n) {
    if (n < 1) throw new IllegalArgumentException();
    if (n==1) System.out.print(1);
    else if (n==2) System.out.print("1 1");
    else {
        System.out.print((n+1)/2+ " ");
        writeSequence(n-2);
        System.out.print(" "+(n+1)/2);
    }
}