factorial.swift

factorial.swift

func factorial(a: Int) -> Int {
    let n = a
    if(n == 1){
      return 1
    }else{
      return n*factorial(n-1)
    }
}

factorial(5)
factorial(6)

func factorial2(a: Int) -> Int {
    return a == 1 ? a : a*factorial(a-1)
}

factorial2(5)
factorial2(6)

let number = 5 //(5!)
var result = 1
for count in 1...number {
result = result * count
}
print(result)

func factorial(of: Int) -> Int
{
return (1...max(of, 1)).reduce(1, *)
}

The formula does not take into account that 0! == 1.

The correct algorithm is:

let n = a
if(n == 0){
return 1
} else {
return n * (n-1).factorial()
}

extension Int {
func fact() -> Int {
return self == 0 || self == 1 ? 1 : self * (self - 1).fact()
}
}

let t = 5.fact()

#!/bin/env swift

extension Int {
	func factorial() -> Self? {
		var tmp = self
		guard tmp > 0 else {return nil}
		if tmp == 0 {
			return 1
		}
		if tmp == 1 {
			return 1
		}
		var result = 1
		while tmp > 1 {
			let multipliedReport = result.multipliedReportingOverflow(by: tmp) // check overflow
			if multipliedReport.overflow { // trigger early return
				return nil
			}
			result = multipliedReport.partialValue
			tmp -= 1
		}
		return result
	}
}

Any Number great than 20, it's factorial will overflow, a.k.a. return nil

func factorial(_ num: UInt) -> UInt {
    guard num != 0 else {
        return 1
    }

    return (1...num).reduce(1, { $0 * $1 })
}

factorial(0) // 1
factorial(5) // 120

Recursion is for hipsters 😜

Note the use of UInt, as factorials are only for positive numbers.

func factorial(_ n: Int) -> Double { return (1...n).map(Double.init).reduce(1.0, *) }