Python FizzBuzz

solution-oop.py

class FizzBuzz:
  def __init__(self,num):
    self.count(num)

  def count(self,num):
    if num >= 1:
      self.count(num - 1)
      self.fizz_buzz(num)

  def fizz_buzz(self,num):
    if num % 15 == 0:
      print "FizzBuzz"
    elif num % 3 == 0:
      print "Fizz"
    elif num % 5 == 0:
      print "Buzz"
    else:
      print num

FizzBuzz(100)

solution-without-loops.py

def count(num):
  if num >= 1:
    count(num - 1)
    fizz_buzz(num)

def fizz_buzz(num):
  if num % 15 == 0:
    print "FizzBuzz"
  elif num % 3 == 0:
    print "Fizz"
  elif num % 5 == 0:
    print "Buzz"
  else:
    print num

count(100)

solution-without-repetition.py

for i in range(1,101):
  string = ""
  if i % 3 == 0:
    string += "Fizz"
  if i % 5 == 0:
    string += "Buzz"
  if i % 5 != 0 and i % 3 != 0:
    string += str(i)
  print string

solution.py

for i in range(1,101):
  if i % 15 == 0:
    print "FizzBuzz"
  elif i % 3 == 0:
    print "Fizz"
  elif i % 5 == 0:
    print "Buzz"
  else:
    print i

print('\n'.join([str(x)*(x%3 != 0 and x%5 != 0) +'Fizz'*(x%3==0)+'Buzz'*(x%5==0) for x in range(1,101)]))

It's disgusting but it's fun

["Fizzbuzz" if i % 15==0 else "Fizz" if i % 3==0 else "Buzz" if i % 5==0 else i for i in range(1,101)]

def fizzbuzz(x):

if x % 3 == 0 and x % 5 == 0:
    return "Fizzbuzz"
elif x % 3 ==0:
    return "Fizz"
elif x % 5 ==0:
    return "Buzz"
else:
    return str(x)

for x in range(1, 101):
print (fizzbuzz(x))

def f(n:int,m:list,w:list):
    s=''
    for x in m:
        if n%x==0:    s+=w[m.index(x)] 
    if s=='':  s=n
    return s
for i in range(1, 100 +1):  print(f(i,[3,5],['Fizz','Buzz']))

i like this solution because it can work with more than just 3 and 5

you can do fizzbuzzfuzz with 3, 5, and 7 just by editing the contents of the lists