问题
def is_prime(x):
if x < 2:
return False
else:
for n in range(2, x):
if x % n == 0:
return False
else:
return True
print is_prime(9) returns True instead of False.
I don't quite understand.
The range (2,9) includes this list: 2,3,4,5,6,7,8
and 9 % 3 == 0,
So how come I do not get False as the answer of that function?
回答1:
This is because you don't actually loop, as you return True during the first cycle (9 % 2 == 0 is False).
Something like this should solve the problem:
def is_prime(x):
if x < 2:
return False
for n in range(2, x):
if x % n == 0:
return False
return True
回答2:
You can simplify the logic a good amount by keeping your original loop and not exiting early. You can add your first conditional to your final return:
def is_prime(x):
for n in range(2, x):
if x % n == 0:
return False
return x > 2
BTW, the Sieve of Erastothenes is a pretty cool method of solving this problem in a much better run time complexity. Here's a link to a brief explanation:
https://math.stackexchange.com/questions/58799/why-in-sieve-of-erastothenes-of-n-number-you-need-to-check-and-cross-out-numbe
来源：https://stackoverflow.com/questions/53244958/how-come-in-this-code-to-find-prime-numbers-is-prime9-returns-true