You are viewing a single comment's thread. Return to all comments →
I've developed this Kotlin algorithm that has O(N) complexity.
fun extraLongFactorials(n: Int): Unit { var result = BigInteger.valueOf(1) (2..n).forEach { value -> result *= BigInteger.valueOf(value.toLong()) } print(result) }
Seems like cookies are disabled on this browser, please enable them to open this website
Extra Long Factorials
You are viewing a single comment's thread. Return to all comments →
I've developed this Kotlin algorithm that has O(N) complexity.