leetcode 326. Power of Three

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326. Power of Three

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Question

Total Accepted: 23021 Total Submissions: 64515 Difficulty: Easy

 

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.

 

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题意：

给一个整数n，判断是否是3的幂

 

class Solution {
public:
    bool isPowerOfThree(int n) {
        double ans = log(n) / log(3);
        double ans2 = floor(ans + 0.5);
        if(fabs(ans - ans2) < 1e-10 ){
            return true;
        }
        else{
            return false;
        }
    }
};

 

改进一下：

class Solution {
public:
    bool isPowerOfThree(int n) {
        double ans = log(n) / log(3);
        double ans2 = round(ans);  //round函数做四舍五入
        if(fabs(ans - ans2) < 1e-10 ){
            return true;
        }
        else{
            return false;
        }
    }
};

 

看了这篇博客的思路，试了一下第三种方法

http://blog.csdn.net/zhoudayang2/article/details/50577721

 

还有要注意边界条件：

class Solution {
public:
    bool isPowerOfThree(int n) {
        if(n < 1){
            return false;
        }
        double ans = log(n) / log(3);
        double ans2 = round(ans);  //round函数做四舍五入
        int m = pow(3,ans2);
        if(n == m){
            return true;
        }
        else{
            return false;
        }
    }
};