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update: 添加问题“326.3的幂”的代码和题解 (#1079)
326: AC.rust+java+go+py+cpp (#1078) rust - AC,51.11%,100.00% java - AC,89.67%,5.37% go - AC,20.00%,41.05% py - AC,46.89%,5.52% cpp - AC,25.79%,19.01% 今天(相比于之前)倒着写的 Signed-off-by: LetMeFly666 <[email protected]>
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‎.commitmsg‎

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close #1074
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326: AC.rust+java+go+py+cpp (#1078)
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rust - AC,51.11%,100.00%
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java - AC,89.67%,5.37%
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go - AC,20.00%,41.05%
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py - AC,46.89%,5.52%
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cpp - AC,25.79%,19.01%
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今天(相比于之前)倒着写的

‎Codes/0326-power-of-three_20250813.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-08-13 13:17:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-08-13 13:25:31
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*/
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class Solution {
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public:
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bool isPowerOfThree(int n) {
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return n > 0 && !(1162261467 % n);
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}
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};

‎Codes/0326-power-of-three_20250813.go‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-08-13 13:17:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-08-13 13:23:55
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*/
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package main
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func isPowerOfThree(n int) bool {
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return n > 0 && 1162261467 % n == 0
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}

‎Codes/0326-power-of-three_20250813.java‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-08-13 13:17:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-08-13 13:23:25
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*/
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class Solution {
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public boolean isPowerOfThree(int n) {
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return n > 0 && 1162261467 % n == 0;
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}
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}

‎Codes/0326-power-of-three_20250813.py‎

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'''
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Author: LetMeFly
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Date: 2025-08-13 13:17:09
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-08-13 13:24:54
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'''
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class Solution:
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def isPowerOfThree(self, n: int) -> bool:
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return n > 0 and not 1162261467 % n

‎Codes/0326-power-of-three_20250813.rs‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-08-13 13:17:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-08-13 13:21:37
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*/
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impl Solution {
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pub fn is_power_of_three(n: i32) -> bool {
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n > 0 && 1162261467 % n == 0
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}
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}

‎Codes/lib.rs‎

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* @LastEditTime: 2025-08-11 22:07:08
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*/
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pub struct Solution;
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include!("2787-ways-to-express-an-integer-as-sum-of-powers.rs"); // 这个fileName是会被脚本替换掉的
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include!("0326-power-of-three_20250813.rs"); // 这个fileName是会被脚本替换掉的

‎README.md‎

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|0316.去除重复字母|中等|<a href="https://leetcode.cn/problems/remove-duplicate-letters/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/09/23/LeetCode%200316.%E5%8E%BB%E9%99%A4%E9%87%8D%E5%A4%8D%E5%AD%97%E6%AF%8D/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127011739" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/remove-duplicate-letters/solution/letmefly-316qu-chu-zhong-fu-zi-mu-dan-di-tczt/" target="_blank">LeetCode题解</a>|
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|0318.最大单词长度乘积|中等|<a href="https://leetcode.cn/problems/maximum-product-of-word-lengths/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/11/06/LeetCode%200318.%E6%9C%80%E5%A4%A7%E5%8D%95%E8%AF%8D%E9%95%BF%E5%BA%A6%E4%B9%98%E7%A7%AF/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134254146" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-product-of-word-lengths/solutions/2515432/letmefly-318zui-da-dan-ci-chang-du-cheng-ytzs/" target="_blank">LeetCode题解</a>|
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|0322.零钱兑换|中等|<a href="https://leetcode.cn/problems/coin-change/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/03/24/LeetCode%200322.%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/136985285" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/coin-change/solutions/2704784/letmefly-322ling-qian-dui-huan-dong-tai-db3o0/" target="_blank">LeetCode题解</a>|
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|0326.3的幂|简单|<a href="https://leetcode.cn/problems/power-of-three/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/08/13/LeetCode%200326.3%E7%9A%84%E5%B9%82/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/150343626" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/power-of-three/solutions/3751661/letmefly-3263-de-mi-da-mi-zheng-chu-by-t-9hdc/" target="_blank">LeetCode题解</a>|
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|0328.奇偶链表|中等|<a href="https://leetcode.cn/problems/odd-even-linked-list/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/09/24/LeetCode%200328.%E5%A5%87%E5%81%B6%E9%93%BE%E8%A1%A8/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127020912" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/odd-even-linked-list/solution/letmefly-328qi-ou-lian-biao-by-tisfy-x9d8/" target="_blank">LeetCode题解</a>|
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|0329.矩阵中的最长递增路径|困难|<a href="https://leetcode.cn/problems/longest-increasing-path-in-a-matrix/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/09/25/LeetCode%200329.%E7%9F%A9%E9%98%B5%E4%B8%AD%E7%9A%84%E6%9C%80%E9%95%BF%E9%80%92%E5%A2%9E%E8%B7%AF%E5%BE%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127043063" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/longest-increasing-path-in-a-matrix/solution/letmefly-329ju-zhen-zhong-de-zui-chang-d-4j3g/" target="_blank">LeetCode题解</a>|
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|0337.打家劫舍III|中等|<a href="https://leetcode.cn/problems/house-robber-iii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/09/27/LeetCode%200337.%E6%89%93%E5%AE%B6%E5%8A%AB%E8%88%8DIII/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127065525" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/house-robber-iii/solution/letmefly-337da-jia-jie-she-iii-by-tisfy-v7ao/" target="_blank">LeetCode题解</a>|

‎Solutions/LeetCode 0326.3的幂.md‎

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---
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title: 326.3 的幂：大幂整除
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date: 2025-08-13 13:26:49
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tags: [题解, LeetCode, 简单, 递归, 数学]
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categories: [题解, LeetCode]
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---
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# 【LetMeFly】326.3 的幂：大幂整除
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力扣题目链接：[https://leetcode.cn/problems/power-of-three/](https://leetcode.cn/problems/power-of-three/)
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<p>给定一个整数，写一个函数来判断它是否是 3&nbsp;的幂次方。如果是，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>
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<p>整数 <code>n</code> 是 3 的幂次方需满足：存在整数 <code>x</code> 使得 <code>n == 3<sup>x</sup></code></p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>n = 27
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<strong>输出：</strong>true
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>n = 0
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<strong>输出：</strong>false
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre>
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<strong>输入：</strong>n = 9
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<strong>输出：</strong>true
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</pre>
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<p><strong>示例 4：</strong></p>
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<pre>
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<strong>输入：</strong>n = 45
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<strong>输出：</strong>false
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>-2<sup>31</sup> &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>进阶：</strong>你能不使用循环或者递归来完成本题吗？</p>
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## 解题方法：看n能否被“最大的”3的幂整除
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题目有一行进阶小字说“你能不使用循环或者递归来完成本题吗？”，好一个挑衅(provocation，bushi)，那就不递归或者循环了吧。
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类似[【LetMeFly】231.2 的幂：五种小方法判断](https://blog.letmefly.xyz/2022/09/08/LeetCode%200231.2%E7%9A%84%E5%B9%82/)的方法五，我们直接使用题目数据范围内最大的3的幂对n取模，看余数是否为0就好了。
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如何求题目数据范围内最大的3的幂？打开python有：
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```python
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>>> 3**20 > 2**31-1
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True
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>>> 3**19 > 2**31-1
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False
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>>> 3**19
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1162261467
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```
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可知如果不大于$2^{31}-1$的正整数$n$是$3$的幂那么它一定能够被$1162261467$整除。
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+ 时间复杂度$O(1)$
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+ 空间复杂度$O(1)$
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### AC代码
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#### C++
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```cpp
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/*
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* @Author: LetMeFly
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* @Date: 2025-08-13 13:17:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-08-13 13:25:31
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*/
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class Solution {
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public:
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bool isPowerOfThree(int n) {
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return n > 0 && !(1162261467 % n);
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}
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};
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```
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#### Python
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```python
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'''
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Author: LetMeFly
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Date: 2025-08-13 13:17:09
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-08-13 13:24:54
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'''
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class Solution:
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def isPowerOfThree(self, n: int) -> bool:
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return n > 0 and not 1162261467 % n
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```
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#### Java
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```java
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/*
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* @Author: LetMeFly
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* @Date: 2025-08-13 13:17:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-08-13 13:23:25
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*/
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class Solution {
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public boolean isPowerOfThree(int n) {
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return n > 0 && 1162261467 % n == 0;
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}
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}
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```
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#### Go
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```go
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/*
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* @Author: LetMeFly
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* @Date: 2025-08-13 13:17:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-08-13 13:23:55
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*/
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package main
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func isPowerOfThree(n int) bool {
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return n > 0 && 1162261467 % n == 0
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}
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```
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#### Rust
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```rust
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/*
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* @Author: LetMeFly
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* @Date: 2025-08-13 13:17:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-08-13 13:21:37
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*/
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impl Solution {
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pub fn is_power_of_three(n: i32) -> bool {
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n > 0 && 1162261467 % n == 0
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}
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}
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```
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> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/150343626)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/08/13/LeetCode%200326.3%E7%9A%84%E5%B9%82/)哦~
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>
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> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)

‎Solutions/Other-English-LearningNotes-SomeWords.md‎

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|shrewd|adj. 机灵的，精明的|
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|prudent|adj. 谨慎的，精打细算的|
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|||
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|provocation|n. 挑衅，刺激，激怒|
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|dignity|n. 尊重，高贵，庄严，端庄，尊严|
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|commonsense|n. 常识，直觉判断力<br/>adj. 常识的，具有常识的|
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+ 这个web要是能设计得可以闭眼(完全不睁眼)键盘控制背单词就好了。
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+ 也许可以加个AI用最近词编故事功能(返回接口中支持标注所使用单词高亮?)

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