326. Power of Three

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?

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题意：

给一个整数n，判断是否是3的幂

 

1 class Solution { 2 public: 3     bool isPowerOfThree(int n) { 4         double ans = log(n) / log(3); 5         double ans2 = floor(ans + 0.5); 6         if(fabs(ans - ans2) < 1e-10 ){ 7             return true; 8         } 9         else{10             return false;11         }12     }13 };

 

改进一下：

1 class Solution { 2 public: 3     bool isPowerOfThree(int n) { 4         double ans = log(n) / log(3); 5         double ans2 = round(ans);  //round函数做四舍五入 6         if(fabs(ans - ans2) < 1e-10 ){ 7             return true; 8         } 9         else{10             return false;11         }12     }13 };

 

看了这篇博客的思路，试了一下第三种方法

 

还有要注意边界条件：

1 class Solution { 2 public: 3     bool isPowerOfThree(int n) { 4         if(n < 1){ 5             return false; 6         } 7         double ans = log(n) / log(3); 8         double ans2 = round(ans);  //round函数做四舍五入 9         int m = pow(3,ans2);10         if(n == m){11             return true;12         }13         else{14             return false;15         }16     }17 };