问题
How would you write a non-recursive algorithm to compute n!?
回答1:
Since an Int32 is going to overflow on anything bigger than 12! anyway, just do:
public int factorial(int n) {
int[] fact = {1, 1, 2, 6, 24, 120, 720, 5040, 40320,
362880, 3628800, 39916800, 479001600};
return fact[n];
}
回答2:
in pseudocode
ans = 1
for i = n down to 2
ans = ans * i
next
回答3:
public double factorial(int n) {
double result = 1;
for(double i = 2; i<=n; ++i) {
result *= i;
}
return result;
}
回答4:
In the interests of science I ran some profiling on various implementations of algorithms to compute factorials. I created iterative, look up table, and recursive implementations of each in C# and C++. I limited the maximum input value to 12 or less, since 13! is greater than 2^32 (the maximum value capable of being held in a 32-bit int). Then, I ran each function 10 million times, cycling through the possible input values (i.e. incrementing i from 0 to 10 million, using i modulo 13 as the input parameter).
Here are the relative run-times for different implementations normalized to the iterative C++ figures:
C++ C#
---------------------
Iterative 1.0 1.6
Lookup .28 1.1
Recursive 2.4 2.6
And, for completeness, here are the relative run-times for implementations using 64-bit integers and allowing input values up to 20:
C++ C#
---------------------
Iterative 1.0 2.9
Lookup .16 .53
Recursive 1.9 3.9
回答5:
Unless you have arbitrary-length integers like in Python, I would store the precomputed values of factorial() in an array of about 20 longs, and use the argument n as the index. The rate of growth of n! is rather high, and computing 20! or 21! you'll get an overflow anyway, even on 64-bit machines.
回答6:
Rewrite the recursive solution as a loop.
回答7:
Here's the precomputed function, except actually correct. As been said, 13! overflows, so there is no point in calculating such a small range of values. 64 bit is larger, but I would expect the range to still be rather reasonable.
int factorial(int i) {
static int factorials[] = {1, 1, 2, 6, 24, 120, 720,
5040, 40320, 362880, 3628800, 39916800, 479001600};
if (i<0 || i>12) {
fprintf(stderr, "Factorial input out of range\n");
exit(EXIT_FAILURE); // You could also return an error code here
}
return factorials[i];
}
Source: http://ctips.pbwiki.com/Factorial
回答8:
long fact(int n) {
long x = 1;
for(int i = 1; i <= n; i++) {
x *= i;
}
return x;
}
回答9:
int total = 1
loop while n > 1
total = total * n
n--
end while
回答10:
I love the pythonic solution to this:
def fact(n): return (reduce(lambda x, y: x * y, xrange(1, n+1)))
回答11:
public int factorialNonRecurse(int n) {
int product = 1;
for (int i = 2; i <= n; i++) {
product *= i;
}
return product;
}
回答12:
At run time this is non-recursive. At compile time it is recursive. Run-time performance should be O(1).
//Note: many compilers have an upper limit on the number of recursive templates allowed.
template <int N>
struct Factorial
{
enum { value = N * Factorial<N - 1>::value };
};
template <>
struct Factorial<0>
{
enum { value = 1 };
};
// Factorial<4>::value == 24
// Factorial<0>::value == 1
void foo()
{
int x = Factorial<4>::value; // == 24
int y = Factorial<0>::value; // == 1
}
回答13:
Pseudo code
total = 1
For i = 1 To n
total *= i
Next
回答14:
fac = 1 ;
for( i = 1 ; i <= n ; i++){
fac = fac * i ;
}
回答15:
assuming you wanted to be able to deal with some really huge numbers, I would code it as follows. This implementation would be for if you wanted a decent amount of speed for common cases (low numbers), but wanted to be able to handle some super hefty calculations. I would consider this the most complete answer in theory. In practice I doubt you would need to compute such large factorials for anything other than a homework problem
#define int MAX_PRECALCFACTORIAL = 13;
public double factorial(int n) {
ASSERT(n>0);
int[MAX_PRECALCFACTORIAL] fact = {1, 1, 2, 6, 24, 120, 720, 5040, 40320,
362880, 3628800, 39916800, 479001600};
if(n < MAX_PRECALCFACTORIAL)
return (double)fact[n];
//else we are at least n big
double total = (float)fact[MAX_PRECALCFACTORIAL-1]
for(int i = MAX_PRECALCFACTORIAL; i <= n; i++)
{
total *= (double)i; //cost of incrimenting a double often equal or more than casting
}
return total;
}
回答16:
I would use memoization. That way you can write the method as a recursive call, and still get most of the benefits of a linear implementation.
回答17:
long fact(int n)
{
long fact=1;
while(n>1)
fact*=n--;
return fact;
}
long fact(int n)
{
for(long fact=1;n>1;n--)
fact*=n;
return fact;
}
回答18:
Iterative:
int answer = 1;
for (int i = 1; i <= n; i++){
answer *= i;
}
Or... using tail recursion in Haskell:
factorial x =
tailFact x 1
where tailFact 0 a = a
tailFact n a = tailFact (n - 1) (n * a)
What tail recursion does in this case is uses an accumulator to avoid piling on stack calls.
Reference: Tail Recursion in Haskell
回答19:
Non recursive factorial in Java. This solution is with custom iterator (to demonstrate iterator use :) ).
/**
* Non recursive factorial. Iterator version,
*/
package factiterator;
import java.math.BigInteger;
import java.util.Iterator;
public class FactIterator
{
public static void main(String[] args)
{
Iterable<BigInteger> fact = new Iterable<BigInteger>()
{
@Override
public Iterator<BigInteger> iterator()
{
return new Iterator<BigInteger>()
{
BigInteger i = BigInteger.ONE;
BigInteger total = BigInteger.ONE;
@Override
public boolean hasNext()
{
return true;
}
@Override
public BigInteger next()
{
total = total.multiply(i);
i = i.add(BigInteger.ONE);
return total;
}
@Override
public void remove()
{
throw new UnsupportedOperationException();
}
};
}
};
int i = 1;
for (BigInteger f : fact)
{
System.out.format("%d! is %s%n", i++, f);
}
}
}
回答20:
For a non-recursive approach, it can't get simpler than this
int fac(int num) {
int f = 1;
for (int i = num; i > 0; i--)
f *= i;
return f;
}
回答21:
int fact(int n){
int r = 1;
for(int i = 1; i <= n; i++) r *= i;
return r;
}
回答22:
Recursively using JavaScript with caching.
var fc = []
function factorial( n ) {
return fc[ n ] || ( ( n - 1 && n != 0 ) &&
( fc[ n ] = n * factorial( n - 1 ) ) ) || 1;
}
来源：https://stackoverflow.com/questions/231250/how-would-you-write-a-non-recursive-algorithm-to-calculate-factorials