[LeetCode]326. Power of Three(判断整数是否是3的幂)

326. Power of Three

原题链接

参考相似题 [LeetCode] 231. Power of Two

Given an integer, write a function to determine if it is a power of three.
给一个整数，写一个函数判断其是否是3的幂

Follow up:

Could you do it without using any loop / recursion?(不用循环和递归)

思路：

• 刚开始想的就是判断边界值，发现3^19 = 1162261467,
• 3^20 > 2147483647(int 最大值)
• 3^0 = 1 (3的最小幂)
• 在1~1162261467范围内所有的3的幂均能被1162261467整除

代码如下：

#include <iostream>using namespace std;class Solution {public:    bool isPowerOfThree(int n) {        if(n<=0 || n>2147483647)//3的幂最小是1  int最大为2147483647            return false;        // 1162261467 is 3^19,  3^20 is bigger than int        return 1162261467%n==0;    }};int main(){    Solution s;    int a =0;    cin >> a;    cout << s.isPowerOfThree(a) << endl;    return 0;}