Am 10.03.2010 16:55, schrieb Bill:
def fizzbuzz(num):
if num:
if num % 15 is 0: return fizzbuzz(num-1) + 'fizzbuzz \n'
elif num % 5 is 0: return fizzbuzz(num-1) + 'buzz \n'
elif num % 3 is 0: return fizzbuzz(num-1) + 'fizz \n'
else : return fizzbuzz(num-1) + ('%d \n' % num)
return ''
print fizzbuzz(100)
While I do understand that this is not a Perl group, and this probably
won't help the OP, I just could not resist:
for i in range(1,101):print('','fizz')[not i%3]+('','buzz')[not i%5]or i
> However, when I try to decipher the logic of the function I imagine
> the solution reversed as "buzz fizz 98 97 ...etc... 4 fizz 2 1".
No, the function creates the string from right to left.
See:
fizzbuzz(100) = fizzbuzz(99) + "buzz \n"
= fizzbuzz(98) + "fizz \n" + "buzz \n"
...
= fizzbuzz(1) + "2 \n" + "fizz \n" + "4 \n" + "buzz\n " + "fizz \n" +
.... + "97 \n" + "98 \n" + "fizz \n" + "buzz \n"
= fizzbuzz(0) + "1 \n" + "2 \n" + "fizz \n" + "4 \n" + "buzz\n " + "fizz
\n" + .... + "97 \n" + "98 \n" + "fizz \n" + "buzz \n"
= " " + "1 \n" + "2 \n" + "fizz \n" + "4 \n" + "buzz\n " + "fizz \n" +
.... + "97 \n" + "98 \n" + "fizz \n" + "buzz \n"
Regards,
Michael
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