Java Fibonacci递归代码

Question

I am trying to come up with a program to take from a user any number and produce the nth number for the fibonacci code. When I complete my work it's showing the next # instead of the # I need. For instance I'm looking for the 11th # and its producing 233 instead of 144. Here is my code:

public static int fibonacci(int n)
{
    if (n<=0)
        return 1;
    else
        return fibonacci(n-2)+ fibonacci(n-1);
}


public static void main(String[] args)
{
    System.out.println("Enter a Number:");
    Scanner keyboard = new Scanner(System.in);
    int number = keyboard.nextInt();
    System.out.println("You Entered Number:" + number);
    System.out.println(number + "th Fibonacci Number is:"+ fibonacci(number));
    keyboard.close();        
}

Answer 1

It's supposed to be

if(n == 0)
    return 0;
else if(n == 1)
    return 1;
else
    return fibonacci(n - 1) + fibonacci(n - 2);

because fibonacci(0) = 0

Answer 2

You are missing an index. You are printing the next Fibonacci number. Change the

(n<=0)

to

(n<=1)

EDIT:

as the other answer says, fib(0) = 0, so you have to add this edge case too.

Answer 3

Aleksandar already gave you the answer, still I think it is necessary to warn you that your code is extremely inefficient. Indeed, for computing fibo(4) , you are going to compute fibo(3) and fibo(2) . fibo(3) will require to compute fibo(2) and fibo(1) etc. Finally, you will take a lot of time computing already computed results (and you will use an exponential amount of memory for storing the local recursive calls context).

For a more efficient code, you ned to implement an iterative or a tail-recursive method. A tail-recursive method meth is a method which recursive calls are of the form meth(f(args)) . In your case, your recursive call has the form f(meth,args) so it is not tail-recursive.

Here is a tail-recursive version :

public static int fibonacci(int n) {
    if (n == 0)
        return 0;
    else if (n <= 2)
        return 1;
    else 
        return fibonacciAux(0,1,n);
}

public static int fibonacciAux(int a, int b, int n) {
    if (count == 0)
        return b;
    else  
        return fibonacciAux(b,a + b,n - 1);
}

This version will run in linear time and will use a constant amount of memory, compared to a non-tail recursive version which will use an exponential time and memory.