Break A Number Up To An Array of Individual Digits

Question

If I have the integer 123 and I want to break the digits into an array [1,2,3] what is the best way of doing this? I have messed around with this a lot and I have the follo

Answer 1

It is easier to work on the UTF-8 representation of the number string because the UTF-8 code unit of a decimal digit can easily be converted to the corresponding integer by subtracting a constant:

let asciiZero = UInt8(ascii: "0")
let digits = map(String(number).utf8) { Int($0 - asciiZero) }

This also turned out to be significantly faster.

If performance is the primary goal then you should restrict the method to simple integer arithmetic, without using strings or characters:

var digits : [Int] = []
while number > 0 {
    digits.insert(number % 10, atIndex: 0)
    number /= 10
}

Here is my complete test code for your convenience (compiled with Xcode 6.4 in Release mode on a MacBook Pro).

func digits1(number : Int) -> [Int] {
    let digits = Array(String(number)).map{Int(strtoul((String($0)), nil, 16))}
    return digits
}

func digits2(number : Int) -> [Int] {
    // Use a static property so that the constant is initialized only once.
    struct Statics {
        static let asciiZero = UInt8(ascii: "0")
    }

    let digits = map(String(number).utf8) { Int($0 - Statics.asciiZero) }
    return digits
}

func digits3(var number : Int) -> [Int] {
    var digits : [Int] = []
    while number > 0 {
        digits.insert(number % 10, atIndex: 0)
        number /= 10
    }
    return digits
}

func measure(converter: (Int)-> [Int]) {
    let start = NSDate()
    for n in 1 ... 1_000_000 {
        let digits = converter(n)
    }
    let end = NSDate()
    println(end.timeIntervalSinceDate(start))
}

measure(digits1) // 10.5 s
measure(digits2) // 1.5 s
measure(digits3) // 0.9 s

---

Update for Swift 3:

func digits(_ number: Int) -> [Int] {
    var number = number
    var digits: [Int] = []
    while number > 0 {
        digits.insert(number % 10, at: 0)
        number /= 10
    }
    return digits
}

print(digits(12345678)) // [1, 2, 3, 4, 5, 6, 7, 8]

This also turned out to be slightly faster than appending the digits to an array and reversing it at the end.