Fibonacci series in Swift

Fibonacci.swift

// Fibonacci series
// F[n] = F[n-1] + F[n-2]
// 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
// Find the fibonacci number for n interations

func fibonacci(n: Int) {

    var num1 = 0
    var num2 = 1

    for _ in 0 ..< n {
    
        let num = num1 + num2
        num1 = num2
        num2 = num
    }
    
    print("result = \(num2)")
}
fibonacci(7)


// Using Recursion
func fibonacciRecursiveNum1(num1: Int, num2: Int, steps: Int) {

    if steps > 0 {
        let newNum = num1 + num2
        fibonacciRecursiveNum1(num2, num2: newNum, steps: steps-1)
    }
    else {
        print("result = \(num2)")
    }
}
fibonacciRecursiveNum1(0, num2: 1, steps: 7)

Why not:
func fibonacci(n: Int) -> Int { if n == 0 || n == 1 { return n; } else { return fibonacci(n: n - 1) + fibonacci(n: n - 2) } }

or

func fibonacci(_ n: Int) -> Int {
    guard n != 0, n != 1 else { return n }
    return fibonacci(n - 1) + fibonacci(n - 2)
}

// returns an array containing the first n Fibonacci numbers
func fibonacci(n: Int) -> [Int] {

    assert(n > 1)

    var array = [0, 1]

    while array.count < n {
        array.append(array[array.count - 1] + array[array.count - 2])
    }
    return array
}

fibonacci(n: 10) // [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

@ALL @leemorgan Need to modify the code from your end what you wrote for Fibonacci

func fibonacci(n: Int) {

var num1 = 0
var num2 = 1

for _ in 0 ..< n {

    let num = num1 + num2
    num1 = num2
    num2 = num
}

print("result = \(num2)")

}

Above is exactly correct for Fibonacci Series of n-1 value

func fibonacci(n: Int) {

    var num1 = 0
    var num2 = 1
    
    for _ in 2...n {
        let num = num1 + num2
        num1 = num2
        num2 = num
    }
    
    print("result = \(num2)")
}

Above is exactly correct for Fibonacci Series of n value

Note::-- Fibonacci Series stats that except 0,1 in it so you need to start for loop from 2 onwards

func fib(_ n: Int) -> Int { return n < 2 ? n : (fib(n-1) + fib(n-2)) }

func fibonacciRecursiveNum1(num1: Int, num2: Int, steps: Int) {

if steps > 0 {
    let newNum = num1 + num2
    fibonacciRecursiveNum1(num2, num2: newNum, steps: steps-1)
}

This is helpful, you can find all kinds of solutions for fibonacci in swift:
https://medium.com/@m.tabrizi/fibonacci-swift-playground-f56d1ff3ea99

here you can find solution with description, I recommend to read:
https://www.weheartswift.com/recursion/

one more quick implementation here:

func fib(_ num: Int) -> Int {
   switch num {
     case Int.min...1: return max(0, num)
     default: return fib(num-2) + fib(num-1)
   }
}

let sequence = Array(0...10)
sequence.forEach { print(fib($0)) } // 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55

careful for future readers, this iterative solution is wrong. f(0) should return 0, f(1) = 1, f(2) = 1

func fibonacci(n: Int) {

    var num1 = 0
    var num2 = 1

    for _ in 0 ..< n {
    
        let num = num1 + num2
        num1 = num2
        num2 = num
    }
    
    print("result = \(num2)")
}
fibonacci(7)

func fibonacci(n : Int){
var n1 = 0
var n2 = 1

if n == 0 {
    
    print("invalid")
    
}else if n == 1 {
    
    print(n1)
    
}else if n == 2 {
    
    print(n1,n2)
    
}else {
    
    var array = [n1,n2]
    
    for _ in 2..<n{
        
        let n3 = n1 + n2
        
        n1=n2
        
        n2=n3
        
        array.append(n3)
        
        
    }
}

}
fibonacci(n: 10)

careful for future readers, this iterative solution is wrong. f(0) should return 0, f(1) = 1, f(2) = 1

func fibonacci(n: Int) {

    var num1 = 0
    var num2 = 1

    for _ in 0 ..< n {
    
        let num = num1 + num2
        num1 = num2
        num2 = num
    }
    
    print("result = \(num2)")
}
fibonacci(7)

THIS CODE IS WRONG

func fibonaccin (n: Int) -> Array{

assert (n > 1)
var array = [0,1]

while array.count < n {
    array.append(array[array.count-1] + array[array.count-2])
    
    print(array)
}
return array

}
fibonaccin(n: 5)

func fib(_ n: Int) -> Int {
    guard n > 1 else { return n }
    return fib(n-1) + fib(n-2)
}

for x in 0...10 {
  print(fib(x))
}

Using decimal you can calculate to big number.

func fibonacci(n: Int) -> [Decimal] {
    
    if n == 0 {
        return [0]
    }
    if n == 1 {
        return [0,1]
    }
    var sequence: [Decimal] = [0,1]
    var num1: Decimal = 0
    var num2: Decimal = 1
    for _ in 2 ..< n+1 {
    
        let num : Decimal = num1 + num2
        num1 = num2
        num2 = num
        sequence.append(num)
    }
    return sequence
}
print("result = \(fibonacci(n: 120))")

This is being a bit clever:

/// Generates a series of `numberOfElements` many fibonacci element, using recursion.
func fibonnaci(numberOfElements n: UInt) -> [Int] {
	let list = [0, 1]
	guard n > 2 else {
		return Array(list.prefix(Int(n)))
	}
	
	func inner(_ l: [Int]) -> [Int] {
		let m = l.count
		guard n != m else { return l }
		return inner(l + [(l[m - 2] + l[m - 1])])
	}
	
	return inner(list)
}

taken from

func fib(_ N: Int) -> Int {
        
    // Init a bottom up array
    var bottomUp: [Int] = []
    
    /**
    Add the 3 known starting cases of 
    the fibonacci sequence
    */
    bottomUp.append(0)
    bottomUp.append(1)
    bottomUp.append(1)
    
    // Check that N is larger than 2
    if N > 2 {
        
        // If so, iterate up through N + 1
        for index in 3...N + 1 {

            // And calculate the fib for that spot
            bottomUp.append(bottomUp[index - 1] + bottomUp[index - 2])
        }
    }
    
    // Return the Nth fib
    return bottomUp[N]
}