[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:

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class Solution {     public boolean isPowerOfThree(int n) {         return n > 0 && Math.pow(3, Math.round(Math.log(n) / Math.log(3))) == n;     } }

Python:

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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:

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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:

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class Solution(object):     def isPowerOfThree(self, n):         """         :type n: int         :rtype: bool         """         return n > 0 and 3 ** round(math.log(n, 3)) == n

Python:

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class Solution(object):     def isPowerOfThree(self, n):         return n > 0 and (math.log10(n)/math.log10(3)).is_integer()

Python:　　

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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++:

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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++:

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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++:

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class Solution { public:     bool isPowerOfThree(int n) {         return n > 0 && pow(3, round(log(n) / log(3))) == n;     } };

　　

　　

 

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