leetcode 326. Power of Three

最新推荐文章于 2025-04-19 00:31:47 发布

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最新推荐文章于 2025-04-19 00:31:47 发布

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原文链接：http://www.cnblogs.com/njczy2010/p/5227703.html

本文介绍了一种算法，用于判断一个整数是否可以表示为3的幂。通过使用对数运算和四舍五入的方法来实现，并考虑了边界条件。

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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

 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

http://blog.youkuaiyun.com/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 };