题意:

  判断是一个数的质因子仅含有2,3,5这3个。

思路:

  因子2比较容易解决,num/=num-(num&num-1)就可以了。3和5的需要通过循环来另判。

C++

 class Solution {
public:
bool isUgly(int num)
{
if(num==) return true;
if(num<) return false;
num/=num-(num&num-);
while(num%==) num/=;
while(num%==) num/=;
return num==;
}
};

AC代码

python3

 class Solution(object):
def isUgly(self, num):
"""
:type num: int
:rtype: bool
"""
if num<1: return False
for x in [2,3,5]:
while num%x==0:
num/=x
return num==1

AC代码

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