我编写了这个函数来返回给定数字的阶乘

func factorial(_ n: Int) -> Int {
    if n == 0 {
        return 1
    }
    else {
        return n * factorial(n - 1)
    }
}

print( factorial(20) )  // 2432902008176640000

只要给定的数字不超过 20，就可以正常工作，因为那样结果就太高了!

我怎样才能规避这个限制，从而计算更高数字的阶乘？

我四处搜索并找到了一些 Swift 的 bignum 库。我这样做是为了学习和熟悉 Swift，因此我想自己解决这个问题。

最佳答案

这是一种可以让您找到非常大的阶乘的方法。

将大数表示为数字数组。例如 987 将是 [9, 8, 7]。将该数字乘以一个整数 n 需要两个步骤。

1. 将该数组中的每个值乘以 n。
2. 执行进位操作以返回个位数的结果。

例如987 * 2:

let arr = [9, 8, 7]
let arr2 = arr.map { $0 * 2 }
print(arr2)  // [18, 16, 14]

现在，执行进位操作。从个位数开始，14 太大了，所以保留4，进位1。将 1 添加到 16 得到 17。

[18, 17, 4]

重复十位:

[19, 7, 4]

然后是百位:

[1, 9, 7, 4]

最后，为了打印，您可以将其转换回字符串:

let arr = [1, 9, 7, 4]
print(arr.map(String.init).joined())

1974

应用该技术，这里有一个执行进位操作的 carryAll 函数，以及一个使用它来计算非常大的阶乘的 factorial:

func carryAll(_ arr: [Int]) -> [Int] {
    var result = [Int]()

    var carry = 0
    for val in arr.reversed() {
        let total = val + carry
        let digit = total % 10
        carry = total / 10
        result.append(digit)
    }

    while carry > 0 {
        let digit = carry % 10
        carry = carry / 10
        result.append(digit)
    }

    return result.reversed()
}



func factorial(_ n: Int) -> String {
    var result = [1]
    for i in 2...n {
        result = result.map { $0 * i }
        result = carryAll(result)
    }

    return result.map(String.init).joined()
}

print(factorial(1000))

402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

关于swift - 在 Swift 3 中，当结果变得太高时如何计算阶乘？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43830151/