Java FizzBuzz recursive solution

Question

Attempting a FizzBuzz recursive solution in Java to return a list of Strings with n iterations. For example, n = 4 should output ["1","2","Fizz", 4]. However, with my current code the output is just ["4"]. Why is my solution not executing the recursive function? Any other critiques are appreciated!

class Solution {
public List<String> fizzBuzz(int n) {

    //create variable to return list of strings
    List<String> fbList = new ArrayList<String>();

    //base case 1
    if(n == 0){
        fbList.add(Integer.toString(0));
    }

    //base case 2
    else if(n == 1){
        fbList.add(Integer.toString(1));
    }    

    //OW take n and begin reducing recursively from, n - 1
    else{
        if(n % 3 == 0){
            fbList.add("Fizz");
        }
        else if(n % 5 == 0){
            fbList.add("Buzz");
        }
        else if((n % 3 == 0) && (n % 5 == 0)){
            fbList.add("FizzBuzz");
        }
        else{
            fbList.add(Integer.toString(n));
        }
        //recursive function call
        fizzBuzz(n - 1);
    }
    return fbList;
    }
}

Answer 1

The problem is that with every recursive call, a new List is created. You return the list but:

fizzBuzz(n - 1);

You are ignoring the return value of the recursive calls. To fix this you can do:

fbList.addAll(0, fizzBuzz(n - 1));

Which will utilize the method addAll to add all the elements returned by the recursive calls. This returns:

[1, 2, Fizz, 4]

However this is rather expensive for an ArrayList. You could change this to a LinkedList, which would allow for linear time adding.

---

Also you if/else if/else chain is out of order. if((n % 3 == 0) && (n % 5 == 0)) should be before if(n % 3 == 0) and if(n % 5 == 0). Otherwise it will always enter the if(n % 5 == 0) or if(n % 3 == 0):

if((n % 3 == 0) && (n % 5 == 0)){
    fbList.add("FizzBuzz");
}
else if(n % 3 == 0){
    fbList.add("Fizz");
}
else if(n % 5 == 0){
    fbList.add("Buzz");
}