Is there a method that calculates a factorial in Java

Question

I didn t find it  yet  Did I miss something   I know a factorial method is a common example program for beginners  But wouldn t it be useful to have a standard implementation for this one to reuse  I could use such a method with standard types  Eg  int  long     and with BigInteger   BigDecimal  too

User · Answer

Factorial is highly increasing discrete function.So I think using BigInteger is better than using int. I have implemented following code for calculation of factorial of non-negative integers.I have used recursion in place of using a loop.

public  BigInteger factorial(BigInteger x){     
    if(x.compareTo(new BigInteger("1"))==0||x.compareTo(new BigInteger("0"))==0)
        return new BigInteger("1");
    else return x.multiply(factorial(x.subtract(new BigInteger("1")))); 
}

Here the range of big integer is

-2^Integer.MAX_VALUE (exclusive) to +2^Integer.MAX_VALUE,
where Integer.MAX_VALUE=2^31.

However the range of the factorial method given above can be extended up to twice by using unsigned BigInteger.

User · Answer

You can use recursion version as well   static int myFactorial int i        if i    1          return      else         System out prinln i    myFactorial --i         Recursion is usually less efficient because of having to push and pop recursions  so iteration is quicker  On the other hand  recursive versions use fewer or no local variables which is advantage

User · Answer

I don t think it would be useful to have a library function for factorial  There is a good deal of research into efficient factorial implementations  Here is a handful of implementations

User · Answer

with recursion   public static int factorial int n        if n    1                return 1                           return n   factorial n-1       with while loop   public static int factorial1 int n        int fact 1      while n gt  1                fact fact n          n--            return fact

User · Answer

i believe this would be the fastest way  by a lookup table   private static final long   FACTORIAL TABLE   initFactorialTable    private static long   initFactorialTable         final long   factorialTable   new long 21       factorialTable 0    1      for  int i 1  i lt factorialTable length  i            factorialTable i    factorialTable i-1    i      return factorialTable           Actually  even for   code long   it works only until 20 inclusively      public static long factorial final int n        if   n  lt  0      n  gt  20           throw new OutOfRangeException  n   0  20       return FACTORIAL TABLE n       For the native type long  8 bytes   it can only hold up to 20   20    2432902008176640000 10    0x 21C3 677C 82B4 0000   Obviously  21  will cause overflow   Therefore  for native type long  only a maximum of 20  is allowed  meaningful  and correct

User · Answer

Bare naked factorials are rarely needed in practice  Most often you will need one of the following   1  divide one factorial by another  or   2  approximated floating-point answer   In both cases  you d be better with simple custom solutions   In case  1   say  if x   90    85   then you ll calculate the result just as x   86   87   88   89   90  without a need to hold 90  in memory     In case  2   google for  Stirling s approximation

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Short answer is  use recursion   You can create one method and call that method right inside the same method recursively   public class factorial        public static void main String   args            System out println calc 10               public static long calc long n            if  n  lt   1              return 1          else             return n   calc n - 1

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Because factorial grows so quickly  stack overflow is not an issue if you use recursion  In fact  the value of 20  is the largest one can represent in a Java long  So the following method will either calculate factorial n  or throw an IllegalArgumentException if n is too big   public long factorial int n        if  n  gt  20  throw new IllegalArgumentException n     is out of range        return  1  gt  n    1   n   factorial n - 1       Another  cooler  way to do the same stuff is to use Java 8 s stream library like this   public long factorial int n        if  n  gt  20  throw new IllegalArgumentException n     is out of range                return LongStream rangeClosed 1  n  reduce 1   a  b  - gt  a   b       Read more on Factorials using Java 8 s streams

User · Answer

I found an amazing trick to find factorials in just half the actual multiplications   Please be patient as this is a little bit of a long post   For Even Numbers   To halve the multiplication with even numbers  you will end up with n 2 factors  The first factor will be the number you are taking the factorial of  then the next will be that number plus that number minus two  The next number will be the previous number plus the lasted added number minus two  You are done when the last number you added was two  i e  2   That probably didn t make much sense  so let me give you an example    8    8    8   6   14     14   4   18     18   2   20   8    8   14   18   20 which is   40320      Note that I started with 8  then the first number I added was 6  then 4  then 2  each number added being two less then the number added before it  This method is equivalent to multiplying the least numbers with the greatest numbers  just with less multiplication  like so   8    1   2   3   4   5   6   7    8     1   8     2   7     3   6     4   5  8    8   14   18   20   Simple isn t it       Now For Odd Numbers  If the number is odd  the adding is the same  as in you subtract two each time  but you stop at three  The number of factors however changes  If you divide the number by two  you will end up with some number ending in  5  The reason is that if we multiply the ends together  that we are left with the middle number  Basically  this can all be solved by solving for a number of factors equal to the number divided by two  rounded up  This probably didn t make much sense either to minds without a mathematical background  so let me do an example    9    9    9   7   16     16   5   21     21   3   24     roundUp 9 2    5   9    9   16   21   24   5     362880     Note  If you don t like this method  you could also just take the factorial of the even number before the odd  eight in this case  and multiply it by the odd number  i e  9    8    9    Now let s implement it in Java   public static int getFactorial int num        int factorial 1      int diffrennceFromActualNum 0      int previousSum num       if num  0    Returning  1 as factorial if number is 0          return 1      if num 2  0     Checking if Number is odd or even                while num-diffrennceFromActualNum gt  2                        if  isFirst                                previousSum previousSum  num-diffrennceFromActualNum                               isFirst false              factorial  previousSum              diffrennceFromActualNum  2                      else    In Odd Case  Number   getFactorial Number-1                 factorial num getFactorial num-1             return factorial      isFirst is a boolean variable declared as static  it is used for the 1st case where we do not want to change the previous sum   Try with even as well as for odd numbers

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public int factorial int num            if  num    1  return 1          return num   factorial num - 1

User · Answer

I got this from EDX use it  its called recursion      public static int factorial int n        if  n    1            return 1        else           return n   factorial n-1

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We need to implement iteratively  If we implement recursively  it will causes StackOverflow if input becomes very big  i e  2 billions   And we need to use unbound size number such as BigInteger to avoid an arithmatic overflow when a factorial number becomes bigger than maximum number of a given type  i e  2 billion for int   You can use int for maximum 14 of factorial and long for maximum 20 of factorial before the overflow   public BigInteger getFactorialIteratively BigInteger input        if  input compareTo BigInteger ZERO   lt   0            throw new IllegalArgumentException  zero or negatives are not allowed               BigInteger result   BigInteger ONE      for  BigInteger i   BigInteger ONE  i compareTo input   lt   0  i   i add BigInteger ONE             result   result multiply i             return result      If you can t use BigInteger  add an error checking   public long getFactorialIteratively long input        if  input  lt   0            throw new IllegalArgumentException  zero or negatives are not allowed          else if  input    1            return 1             long prev   1      long result   0      for  long i   2  i  lt   input  i              result   prev   i          if  result   prev    i       check if result holds the definition of factorial                arithmatic overflow  error out             throw new RuntimeException  value   i   is too big to calculate a factorial  prev   prev    current   result                     prev   result            return result

User · Answer

Try this  public static BigInteger factorial int value       if value  lt  0           throw new IllegalArgumentException  Value must be positive               BigInteger result   BigInteger ONE      for  int i   2  i  lt   value  i              result   result multiply BigInteger valueOf i               return result

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A fairly simple method      for   int i   1  i  lt  n   i                       answer   answer   i

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The only business use for a factorial that I can think of is the Erlang B and Erlang C formulas  and not everyone works in a call center or for the phone company  A feature s usefulness for business seems to often dictate what shows up in a language - look at all the data handling  XML  and web functions in the major languages   It is easy to keep a factorial snippet or library function for something like this around

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Apache Commons Math has a few factorial methods in the MathUtils class

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public static long factorial int number        if  number  lt  0            throw new ArithmeticException number    quot  is negative quot              long fact   1      for  int i   1  i  lt   number    i            fact    i            return fact      using recursion   public static long factorial int number        if  number  lt  0            throw new ArithmeticException number    quot  is negative quot              return number    0    number    1   1   number   factorial number - 1       source

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A very simple method to calculate factorials   private double FACT double n        double num   n      double total   1      if num    0   num    1           total   num       else if num    1   num    0           total   1            double num2      while num  gt  1           num2   num - 1          total   total   num2          num   num - 1            return total      I have used double because they can hold massive numbers  but you can use any other type like int  long  float  etc   P S  This might not be the best solution but I am new to coding and it took me ages to find a simple code that could calculate factorials so I had to write the method myself but I am putting this on here so it helps other people like me

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import java liberary class     import java util Scanner      class to find factorial of a number     public class factorial   public static void main String   args        scanner method for read keayboard values      Scanner factor  new Scanner System in        int n      double total   1      double sum  1       System out println   nPlease enter an integer          n   factor nextInt        evaluvate the integer is greater than zero and calculate factorial  if n  0         System out println   Factorial of 0 is 1      else if  n gt 0        System out println   nThe factorial of     n     is           System out print n        for int i 1 i lt n i                  do    do while loop for display each integer in the factorial                                 System out print      n-i                              while   n    1          total   total   i             calculate factorial sum  total   n       display sum of factorial      System out println   n nThe    n    Factorial is          sum         display invalid entry  if enter a value less than zero  else        System out println   nInvalid entry        System exit 0

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public class UsefulMethods       public static long factorial int number            long result   1           for  int factor   2  factor  lt   number  factor                  result    factor                     return result            Big Numbers version by HoldOffHunger   public static BigInteger factorial BigInteger number        BigInteger result   BigInteger valueOf 1        for  long factor   2  factor  lt   number longValue    factor              result   result multiply BigInteger valueOf factor               return result

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public static int fact int i       if i  0         return 0      if i gt 1          i   i   fact --i             return i

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USING DYNAMIC PROGRAMMING IS EFFICIENT   if you want to use it to calculate again and again  like caching   Java code   int fact   new int n 1     n is the required number you want to find factorial for  int factorial int num         if num  0        fact num  1       return fact num                 else        fact num   num  factorial num-1         return fact num

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Although factorials make a nice exercise for the beginning programmer  they re not very useful in most cases  and everyone knows how to write a factorial function  so they re typically not in the average library

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using recursion is the simplest method  if we want to find the factorial of N  we have to consider the two cases where N   1 and N gt 1 since in factorial we keep multiplying N N-1  N-2      until 1   if we go to N  0 we will get 0 for the answer  in order to stop the factorial reaching zero  the following recursive method is used  Inside the factorial function while N gt 1  the return value is multiplied with another initiation of the factorial function  this will keep the code recursively calling the factorial   until it reaches the N  1  for the N 1 case  it returns N  1  itself and all the previously built up result of multiplied return N s gets multiplied with N 1  Thus gives the factorial result  static int factorial int N        if N  gt  1         return n   factorial N - 1                Base Case N   1     else        return N

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We have a single line to calculate it   Long factorialNumber   LongStream rangeClosed 2  N  reduce 1  Math  multiplyExact

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You can use recursion   public static int factorial int n             if  n    0              return 1            else             return n   factorial n-1                 and then after you create the method function  above   System out println factorial number of your choice            direct example     System out println factorial 3

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Apache Commons Math package has a factorial method  I think you could use that

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while loop  for small numbers   public class factorial    public static void main String   args        int counter 1  sum 1       while  counter lt  10            sum sum counter          counter              System out println  Factorial of 10 is    sum

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Use Guava s BigIntegerMath as follows   BigInteger factorial   BigIntegerMath factorial n      Similar functionality for int and long is available in IntMath and LongMath respectively

[java] Is there a method that calculates a factorial in Java?

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