问题
how can I combine these two functions in to one recursive function to have this result:
factorial(6)
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
these are the codes
def factorial( n ):
if n <1: # base case
return 1
else:
return n * factorial( n - 1 ) # recursive call
def fact(n):
for i in range(1, n+1 ):
print "%2d! = %d" % ( i, factorial( i ) )
fact(6)
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
as you see execution of these two gives a correct answer, I just want to make it to one recursive function.
回答1:
def factorial( n ):
if n <1: # base case
return 1
else:
returnNumber = n * factorial( n - 1 ) # recursive call
print(str(n) + '! = ' + str(returnNumber))
return returnNumber
回答2:
2 lines of code:
def fac(n):
return 1 if (n < 1) else n * fac(n-1)
Test it:
print fac(4)
Result:
24
回答3:
a short one:
def fac(n):
if n == 0:
return 1
else:
return n * fac(n-1)
print fac(0)
回答4:
def factorial(n):
result = 1 if n <= 1 else n * factorial(n - 1)
print '%d! = %d' % (n, result)
return result
回答5:
try this:
def factorial( n ):
if n <1: # base case
print "%2d! = %d" % (n, n)
return 1
else:
temp = factorial( n - 1 )
print "%2d! = %d" % (n, n*temp)
return n * temp # recursive call
One thing I noticed is that you are returning '1' for n<1, that means your function will return 1 even for negative numbers. You may want to fix that.
回答6:
I've no experience with Python, but something like this?
def factorial( n ):
if n <1: # base case
return 1
else:
f = n * factorial( n - 1 ) # recursive call
print "%2d! = %d" % ( n, f )
return f
回答7:
fac = lambda x: 1 if x == 0 else x * fac(x - 1)
回答8:
Is this homework by any chance?
def traced_factorial(n):
def factorial(n):
if n <= 1:
return 1
return n * factorial(n - 1)
for i in range(1, n + 1):
print '%2d! = %d' %(i, factorial(i))
Give PEP227 a read for more details. The short of it is that Python lets you define functions within functions.
回答9:
One more
def fact(x):
if x == 0:
return 0
elif x == 1:
return 1
else:
return x * fact(x-1)
for x in range(0,10):
print '%d! = %d' %(x, fact(x))
回答10:
I don't really know the factorial of negative numbers, but this will work with all n >= 0:
def factorial(n):
if n >= 0:
if n == 1 or n==0:
return 1
else:
n = n * factorial(n-1)
return n
else:
return 'error'
回答11:
Can use this 4 lines of code...
def factorial(n):
f = lambda n: n * f(n - 1) if n > 1 else 1
for x in range(n):
print('{}! = {}'.format(x + 1, factorial(x + 1)))
回答12:
And for the first time calculate the factorial using recursive and the while loop.
def factorial(n):
while n >= 1:
return n * factorial(n - 1)
return 1
Although the option that TrebledJ wrote in the comments about using if
is better. Because while
loop performs more operations (SETUP_LOOP, POP_BLOCK
) than if
. The function is slower.
def factorial(n):
if n >= 1:
return n * factorial(n - 1)
return 1
timeit -n 10000 -r 10
while
836 µs ± 11.8 µs per loopif
787 µs ± 7.22 µs per loop
回答13:
There is always some kind of a loop in recursive functions and some stoping codes which stop the loop:
public int recursivefactorial(int number)
{
if(number==1)
return 1;
else
return recursivefactorial(number-1)*number;
}
As you can see the fulfilling the if
condition leads to the code that actually ends the "loop" and this is the most important part of a recursive function. In contrast, the else
part of the condition leads to calling recursivefactorial
function once again which is effectively a kind of loop.
回答14:
One more =)
#FAC calculation
def fakulteta(x):
if x!=1:
return x*fakulteta(x-1)
return 1
print (fakulteta(12))
来源:https://stackoverflow.com/questions/4502429/recursive-factorial-function