swift 中的递归斐波那契 function

Question

I tried to write recursive function for Fibonacci sequence. I used an array to save precalculated elements to improve algorithm(which is a common way to do).

Here is my code:

var n = Int(readLine()!)

var arr = [Int](repeating:0, count:n!)
arr[0] = 1
arr[1] = 1
arr[2] = 2


func fib(n : Int, arr: inout [Int]) -> (Int,[Int]){

    if arr[0] != 0 {
        return (arr[n],arr)
    }

    arr[n] = fib(n: n - 1,arr: &arr).0 + fib(n: n - 2,arr: &arr).0
    return (arr[n],arr)
}
n! -= 1
print(fib(n:n! ,arr: &arr).0)

Notice: 3 < n

The answer for any integer as n is 0.

How can I fix it?

I know that using global variables are much easier(for saving operation), but i don't know how to do that.

Answer 1

The mistake seems to be in the first if statement:

if arr[0] != 0 {
    return (arr[n], arr)
}

If the first element of the array is not 0, then you return arr[n] . Well, arr[0] is always 1, so the condition is always true, but arr[n] is initially 0 for all n >= 3, which is what caused the observed behaviour.

I think you meant to say:

if arr[n] != 0 {

Also, the function doesn't need to return the array since you are using inout - passing the array by reference. You are not using the second item of the tuple anywhere, are you? So the function can be written as:

func fib(n : Int, arr: inout [Int]) -> Int {

    if arr[n] != 0 {
        return arr[n]
    }

    arr[n] = fib(n: n - 1,arr: &arr) + fib(n: n - 2,arr: &arr)
    return arr[n]
}

Answer 2

func fibonacciOf(_ number: Int) -> Int {
    if number == 1 || number == 2 {
        return 1
    }

    return fibonacciOf(number - 2) + fibonacciOf(number - 1)
}

print(fibonacciOf(5))