[LeetCode] 326. Power of Three 3的次方数
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?
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
给一个整数，写一个函数来判断此数是不是3的次方数。
类似的题目Power of Two 中，由于2的次方数的特点，用位操作很容易。而3的次方数没有显著的特点，最直接的方法就是不停地除以3，最后判断是否能整除。
follow up是否不用任何循环或递归。
解法1: 循环
解法2: 迭代
解法3：取对数
Java:
class Solution {
public boolean isPowerOfThree(int n) {
return n > 0 && Math.pow(3, Math.round(Math.log(n) / Math.log(3))) == n;
}
}
Python:
class Solution(object):
def isPowerOfThree(self, n):
"""
:type n: int
:rtype: bool
"""
if n <= 0: return False
while n != 1:
if n % 3 != 0: return False
n /= 3
return True
Python:
class Solution(object):
def isPowerOfThree(self, n):
"""
:type n: int
:rtype: bool
"""
if n <= 0: return False
if n == 1: return True
return n % 3 == 0 and self.isPowerOfThree(n / 3)
Python:
class Solution(object):
def isPowerOfThree(self, n):
"""
:type n: int
:rtype: bool
"""
return n > 0 and 3 ** round(math.log(n, 3)) == n
Python:
class Solution(object):
def isPowerOfThree(self, n):
return n > 0 and (math.log10(n)/math.log10(3)).is_integer()
Python:
class Solution(object):
def __init__(self):
self.__max_log3 = int(math.log(0x7fffffff) / math.log(3))
self.__max_pow3 = 3 ** self.__max_log3
def isPowerOfThree(self, n):
"""
:type n: int
:rtype: bool
"""
return n > 0 and self.__max_pow3 % n == 0
C++:
class Solution {
public:
bool isPowerOfThree(int n) {
if(n <= 0) return false;
while(n > 1){
if(n %3 != 0) return false;
n/=3;
}
return true;
}
};
C++:
class Solution {
public:
bool isPowerOfThree(int n) {
if (n <= 0) return false;
if (n == 1) return true;
return n % 3 == 0 && isPowerOfThree(n / 3);
}
};
C++:
class Solution {
public:
bool isPowerOfThree(int n) {
return n > 0 && pow(3, round(log(n) / log(3))) == n;
}
};
