leetcode Power of Three

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?

题目地址： leetcode Power of Three

题目大意：

给定一个整数，判断是否是3的整数幂

思路

方法一

循环判断，每次除以3

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;
    }
};

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

方法二

递归

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);
	}
};

 

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)

 

方法三

取对数

C++

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

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
        """
        return n > 0 and 3 ** round(math.log(n, 3)) == n

 

本文是leetcode 326 Power of Three 题解，更多本博客题解见 https://www.hrwhisper.me/leetcode-algorithm-solution/