我已编写此函数以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,因为这样结果会变得太高!
我如何规避此限制,从而计算出较高数字的阶乘?
我四处搜寻,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。
这里是一种可以让您找到非常大的阶乘的方法。
将大数字表示为数字数组。例如,987
将为[9, 8, 7]
。将该数字乘以整数n
将需要两个步骤。
n
。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。从一个数字开始,14
太大,因此请保留4
并随身携带1
。将1
添加到16
,以获得17
。
[18, 17, 4]
用十位代替:
[19, 7, 4]
然后是百位:
[1, 9, 7, 4]
最后,对于打印,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,这是执行进位运算的carryAll
函数,并使用它来计算非常大的阶乘: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))
40238726007709377354370243392300398571937486421071463254379991042993851239862902059204420848696940480047998861019719605863166687299480855890132382966994459099742450408707375991882362772718873251977950595099527612087497546249704360141827809464649629105639388743788648733711918104582578364784997701247663288983595573543251318532395846307555740911426241747434934755342864657661166779739666882029120737914385371958824980812686783837455973174613608537953452422158659320192809087829730843139284440328123155861103697680135730421616874760967587134831202547858932076716913244842623613141250878020800026168315102734182797770478463586817016436502415369139828126481021309276124489635992870511496497541990934222156683257208082133318611681155361583654698404670897560290095053761647584772842188967964624494516076535340819890138544248798495995331910172335555660213945039973628075013783761530712776192684903435262520001588853514733161170210396817592151090778801939317811419454525722386554146106289218796022383897147 6088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863 554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
您可以使用此库:BigInt
使用CocoaPods安装它:
pod 'BigInt'
然后您可以像这样使用它:
import BigInt
func factorial(_ n: Int) -> BigInt {
if n == 0 {
return 1
}
else {
return BigInt(n) * factorial(n - 1)
}
}
print( factorial(50) ) // 30414093201713378043612608166064768844377641568960512000000000000