412 fizz buzz
412. Fizz Buzz
class Solution(object): def fizzBuzz(self, n): """ :type n: int :rtype: List[str] """ return [(not i%3)*"Fizz" + (not i%5)*"Buzz" or str(i) for i in range(1, n+1)]
class Solution(object): def fizzBuzz(self, n): """ :type n: int :rtype: List[str] """ return [str(i) if (i%3!=0 and i%5!=0) else (('Fizz'*(i%3==0)) + ('Buzz'*(i%5==0))) for i in range(1,n+1)]
# the fizbuz logic, returns an iterator object that # calculates one value at a time, not all ot them at once def fiz(numbers): for i in numbers: if i % 15 == 0: yield 'fizbuz' elif i % 5 == 0: yield 'buz' elif i % 3 == 0: yield 'fiz' else: yield str(i) # xrange evaluates lazily, good for big numbers # matches well with the lazy-eval generator function numbers = xrange(1,2**20) # this gets one number, turns that one number into fuz, repeat print ' '.join(fiz(numbers)) # returns: 1 2 fiz 4 buz fiz [...] fiz 1048573 1048574 fizbuz
- clearly separates fizbuz logic from concatenation
- is as plain and readeable as possible
- generator iterator does not keep all the array in memory
- so that you can do it on arbitrary numbers (see Euler problem #10)
What I do not like in this solution is the three ifs, whereas the problem can be solved with two.
Answer: because yield is efficient when you do not want to keep big arrays in memory just to iterate through them. But this question is not about big arrays.