[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;
}