This is an easy problem. What we need to do is to check if a number i in ranges from 1 to n has the divisor 3 and 5.
There are many ways to realize it in Python.

 Use a loop containing condition languages. This is simple and understood by everyone. And below is my solution.
class Solution(object):
def fizzBuzz(self, n):
"""
:type n: int
:rtype: List[str]
"""
ans=[]
for i in range(1,n+1):
if i % 15==0:
ans.append("FizzBuzz")
elif i % 3==0:
ans.append("Fizz")
elif i % 5==0:
ans.append("Buzz")
else:
ans.append(str(i))
return ans

 I also see some amazing short solutions in the Discuss forum. These guys are really masters! Below are solutions provided by @StefanPochmann. He only writes one line. These short solutions cannot be realized without the convenience and flexibility of Python.
 2.1.
def fizzBuzz(self, n):
return['FizzBuzz'[i%3&4:i%5&8^12]or`i`for i in range(1,n+1)]
 2.2.
def fizzBuzz(self, n):
return ['Fizz' * (not i % 3) + 'Buzz' * (not i % 5) or str(i) for i in range(1, n+1)]