【LeetCode】326. Power of Three

LeetCode-326

Links:https://leetcode.com/problems/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?

思路：

1.用2^31次方内，最大的3的幂去除n取余…判断余数是否为0

2.用log()函数.判断n是3的多少次方（是不是一个整数）,用log10()是因为精度问题.

代码1：

class Solution {
public:
    bool isPowerOfThree(int n) {
        return n>0 && 1162261467%n==0;
    }
};

class Solution {

public:

bool isPowerOfThree(int n) {

return n>0 && 1162261467%n==0;

}

};

代码2：

class Solution {
public:
    bool isPowerOfThree(int n) {
        double asnlog = log10(n)/log10(3);
        return asnlog- int(asnlog)==0?true:false;
    }
};

class Solution {

public:

bool isPowerOfThree(int n) {

double asnlog = log10(n)/log10(3);

return asnlog- int(asnlog)==0?true:false;

}

};

【LeetCode】326. Power of Three

Tagged on: LeetCode