recursion problem

[katiebear128]

public static void recur2 (int n)
{
  if(n<=0)
  { 
    return;
  }
  else
  {
    recur2(n - 2);
    System.out.print(n);
  }
}

My problem: What does recur2(5) display?

I tried putting the method into a class w/ a main method like so:

public class recur2{
  
  public static void main(String[] args){

  }//end main
  
  public static void recur2 (int n)
{
  if(n<=0)
  { 
    return;
  }
  else
  {
   recur2(n - 2);
   System.out.print(n);
  }

}

}

But I can't put System.out.println(recur(5)) anywhere without it telling me a void statement can't go there...How can I get it to run?

But thinking it through, I know for 5 it go to the else statement first: so you get recur2(5-2);
So it prints out 5, but in the background there is 3! still going on. takes the 3! goes to else statement again get 1!, but it prints out 3. Is that logic at all right? And I don't get what the base case is...it does says return. return what?

[sunde887]

You don't need to print the method, just call it.

public static void main(String[] args){
  //call method
}

[katiebear128]

thanks for pointing that out! I got it :)