Kotlin algorithm Getting started calculating primes and optimizing __ algorithms

classGetprimenumber {/* * gets [1,n] interval primes */ FunForeachnumbergetprime (number:int): list<int> {Valintegers = arraylist<int> () for(Iinch1 until number)if(Isprimenumber (2, I)) Integers.add (i)returnintegers}/** * Gets the prime number from the specified interval */ FunForeachnumbergetprimetospecifiedposition (Startposition:int, Endposition:int): list<int> {Valintegers = arraylist<int> () for(IinchStartPosition until Endposition)if(Isprimenumber (2, I)) Integers.add (i)returnintegers}/** * Print a prime number from 1 to n */ FunForeachprintnumbergetprime (number:int) { for(Iinch1 until number)if(Isprimenumber (2, I)) println ("number = [$I] is the prime number ")}/** * Print the prime number from the specified interval */ FunForeachprintnumbergetprimetospecifiedposition (Startposition:int, Endposition:int) { for(IinchStartPosition until Endposition)if(Isprimenumber (2, I)) println ("number = [$I] is the prime number ")}/** * Because the non-prime number can be calculated by multiplying by 1 9 in addition to 1 and 2 only if it is necessary to continue to be divisible by 2-9 * This statement uses the extraction of the minimum common divisor to calculate * Of course to avoid a heavy The problem is that when it is a single digit, it is 1, 2, 3, 5, 7, so the direct return of 20 of the prime number * The advantage of this calculation is to avoid the traditional recursion from 1 to n the repeated calculation of more efficient calculation of prime numbers facing more than thousands of data use * Also avoids excessive use of this Algorithm (redundant repeatability calculation): The method of judging the prime number: to remove 2 to sqrt (the square root of this number), if it can be divisible, it means that the number is not a prime, the inverse is prime, the algorithm is faster * Avoid redundant redundancy */ FunIsprimenumber (Divisor:int, Number:int): Boolean {if(number% Divisor = = 0)return Falseelse if(number = = 1 | | number = = 2 | | number = = 3 | | number = = 5 | | number = = 7 | | | Number = = 17 | | Number = = 19)return Trueelse if(Number <= 20)return Falseelse if(divisor = = 9) {returnIsprimenumber (one, divisor)}else if(Divisor > 9) {return if(Divisor < Math.sqrt (number.todouble ())) {Isprimenumber (divisor + 1, number)}else if(divisor.todouble () = = Math.sqrt (number.todouble ()))falseElsetrue }returnIsprimenumber (divisor + 1, number)}}