LeetCode-power_of_three
Given an integer, write a function to determine if it is a power of three.
Follow up:
Could you do it without using any loop / recursion?
# _*_ coding: utf-8 _*_
class Solution(object):
def isPowerOfThree(self, n):
"""
:type n: int
:rtype: bool
"""
import math
if n < 1:
return False
# 四舍五入, 因为最后的取对数会有少许误差
return 3 ** round(math.log(n, 3)) == n
def isPowerOfThree1(self, n):
"""
:type n: int
:rtype: bool
思路是在看了别人的解析后得出:
直接在32位的有符号整型号里，找到最大的3的幂数，然后用n对它进行取余即可
In [178]: math.log(1162261467, 2)
Out[178]: 30.11428751370197
In [180]: math.log(1162261467 * 3, 2)
Out[180]: 31.699250014423125
32位最大3的幂数是1162261467
"""
return n > 0 and 1162261467 % n == 0