leetcode 326. Power of Three

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326. Power of Three

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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，判断是不是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函数作四舍五入
 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

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

 

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函数作四舍五入
 9         int m = pow(3,ans2);
10         if(n == m){
11             return true;
12         }
13         else{
14             return false;
15         }
16     }
17 };