update: 添加问题“342.4的幂”的代码(并更新其题解) (#1083)

* 342: RE(+WA).cpp (#1082)

* 342: AC.cpp (#1082)

cpp - AC,100.00%,82.09%

* 342: AC.cpp (#1082) - mod 3

cpp - AC,100.00%,33.86%

* 342: py.cpp (#1082) - mod 3

py - AC,12.55%,65.26%

* 342: AC.go+WA.java (#1082) - mod 3

go - AC,100.00%,42.17%

* 342: AC.java (#1082) - mod 3

java - AC,100.00%,40.06%

* 342: AC.rust (#1082) - mod 3

rust - AC,100.00%,33.33%

* update: 添加问题“342.4的幂”的代码(并更新其题解) (#1083)
326: AC.rust+java+go+py+cpp (#1078)

cpp - AC,100.00%,43.53%
py - AC,100.00%,59.18%
go - AC,100.00%,90.00%
java - AC,100.00%,75.56%
rust - AC,100.00%,-%  # 是的，没看错

Signed-off-by: LetMeFly666 <[email protected]>

---------

Signed-off-by: LetMeFly666 <[email protected]>

Author: LetMeFly666

Codes/0342-power-of-four_20250815-sqrt.cpp

@@ -0,0 +1,35 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-15 18:29:00
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-15 18:50:45
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        if (n <= 0) {
+            return false;
+        }
+        int k = sqrt(n);
+        return k * k == n && (n & (n - 1)) == 0;
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+16
+
+*/
+int main() {
+    int n;
+    while (cin >> n) {
+        Solution sol;
+        cout << sol.isPowerOfFour(n) << endl;
+    }
+    return 0;
+}
+#endif

Codes/0342-power-of-four_20250815.cpp

@@ -0,0 +1,16 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-15 18:29:00
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-15 18:49:20
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && n % 3 == 1;
+    }
+};

Codes/0342-power-of-four_20250815.go

@@ -0,0 +1,11 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-15 18:29:00
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-15 23:54:26
+ */
+package main
+
+func isPowerOfFour(n int) bool {
+    return n > 0 && n & (n - 1) == 0 && n % 3 == 1
+}

Codes/0342-power-of-four_20250815.java

@@ -0,0 +1,11 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-15 18:29:00
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-15 23:56:02
+ */
+class Solution {
+    public boolean isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && n % 3 == 1;
+    }
+}

Codes/0342-power-of-four_20250815.py

@@ -0,0 +1,9 @@
+'''
+Author: LetMeFly
+Date: 2025-08-15 18:29:00
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-08-15 18:52:04
+'''
+class Solution:
+    def isPowerOfFour(self, n: int) -> bool:
+        return n > 0 and n & (n - 1) == 0 and n % 3 == 1

Codes/0342-power-of-four_20250815.rs

@@ -0,0 +1,11 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-15 18:29:00
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-15 23:57:22
+ */
+impl Solution {
+    pub fn is_power_of_four(n: i32) -> bool {
+        n > 0 && n & (n - 1) == 0 && n % 3 == 1
+    }
+}

Codes/lib.rs

@@ -5,4 +5,4 @@
  * @LastEditTime: 2025-08-11 22:07:08
  */
 pub struct Solution;
-include!("1780-check-if-number-is-a-sum-of-powers-of-three_20250814.rs");  // 这个fileName是会被脚本替换掉的
+include!("0342-power-of-four_20250815.rs");  // 这个fileName是会被脚本替换掉的

Solutions/LeetCode 0342.4的幂.md

@@ -81,5 +81,94 @@ public:
 };
 ```
 
-> 同步发文于CSDN，原创不易，转载请附上[原文链接](https://blog.letmefly.xyz/2022/09/29/LeetCode%200342.4%E7%9A%84%E5%B9%82/)哦~
-> Tisfy：[https://letmefly.blog.csdn.net/article/details/127104821](https://letmefly.blog.csdn.net/article/details/127104821)
+## 方法二：判断是否 是2的幂且是平方数
+
+如何判断一个数是否是2的幂？这里有5种方法判断：[【LetMeFly】231.2 的幂：五种小方法判断](http://blog.letmefly.xyz/2022/09/08/LeetCode%200231.2%E7%9A%84%E5%B9%82/)
+
+如果一个数是2的幂，那么它是4的幂吗？只需要看这个数是否是平方数即可（$\lfloor\sqrt{n}\rfloor^2==n$）。
+
+这是因为$4^t=(2^{\frac{t}{2}})^2$。
+
++ 时间复杂度$O(1)$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        if (n <= 0) {
+            return false;
+        }
+        int k = sqrt(n);
+        return k * k == n && (n & (n - 1)) == 0;
+    }
+};
+```
+
+## 方法三：判断是否 是2的幂且模3得1
+
+如果一个数是2的幂，那么它是4的幂等价于它模3得1。
+
++ 时间复杂度$O(1)$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && n % 3 == 1;
+    }
+};
+```
+
+#### Python
+
+```python
+class Solution:
+    def isPowerOfFour(self, n: int) -> bool:
+        return n > 0 and n & (n - 1) == 0 and n % 3 == 1
+```
+
+#### Java
+
+```java
+class Solution {
+    public boolean isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && n % 3 == 1;
+    }
+}
+```
+
+#### Go
+
+```go
+package main
+
+func isPowerOfFour(n int) bool {
+    return n > 0 && n & (n - 1) == 0 && n % 3 == 1
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn is_power_of_four(n: i32) -> bool {
+        n > 0 && n & (n - 1) == 0 && n % 3 == 1
+    }
+}
+```
+
+## End
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/127104821)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/2022/09/29/LeetCode%200342.4%E7%9A%84%E5%B9%82/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)

Solutions/Other-English-LearningNotes-SomeWords.md

@@ -1073,7 +1073,7 @@ categories: [自用]
 |detract|v. 破坏，损害，转移，使分心<details><summary>例句</summary>Too much make-up will <font color="#28bea0">detract</font> from her beauty.<br/>画太浓的妆有损她的美丽。</details>|
 |esteem|n. 尊重，敬重，好评<br/>v. 尊重，敬重，认为，把...看作|
 |||
-|piston|n. 活塞|
+|<font color="#28bea0" title="二次复习">piston</font>|n. 活塞|
 |||
 |postponement|n. 延期，推迟|
 |||
@@ -1360,6 +1360,9 @@ categories: [自用]
 |commonsense|n. 常识，直觉判断力<br/>adj. 常识的，具有常识的|
 |||
 |solemn|adj. 严肃的，庄严的，正式的，郑重的|
+|||
+|conscientiously|adv. 完全地，彻底地，负责地|
+|headmaster|n. (中小学)校长|
 
 + 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。
 + 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?)