write a function that takes an int as it parameter and returns an array of longs where each element has the value of the factorial of it's index.

Question

f(n) --> [1!,2!,3!, ..., (n-1)!,n!]

Answer 1

import java.util.Scanner;

class JeffC

{

private Scanner scanner;

public static final int MAX_FACTORIAL = 20;

JeffC()

{

scanner = new Scanner(System.in);

}

public long factorial( int n)

{

long longFactorialReturn = -1;

if (n>=0)

{

if ((n==0) || (n==1))

{

longFactorialReturn = 1;

}

else

{

longFactorialReturn=n * factorial(n-1);

}

}

return(longFactorialReturn);

}

public long[] Factorials ( int n)

{

long factorialArrayReturn[] = new long[n+2]; //0!, 1!,2!, .... (n-1)!, n! is (n+1) long values

for (int iLoop=0; iLoop<=n; iLoop++)

{

factorialArrayReturn[iLoop]=factorial(iLoop);

}

return(factorialArrayReturn);

}

public void Go()

{

int nInput = -1;

while ((nInput<0) || (nInput>MAX_FACTORIAL))

{

System.out.print("Please input the integer value :>");

nInput = scanner.nextInt();

if (nInput<0) { System.out.println("\n Say what? \n"); }

if (nInput>MAX_FACTORIAL) { System.out.println(" \n No, sorry too many "); }

}

long factorials[] = Factorials(nInput);

for (int iLoop=0; iLoop<=nInput; iLoop++)

{

System.out.println( iLoop + "! = " + factorials[iLoop] );

}

}

public static void main(String args[])

{

JeffC x = new JeffC();

x.Go();

}

}