feat: add solutions to lc problems: No.0231,0342

Author: yanglbme

solution/0200-0299/0231.Power of Two/README.md

@@ -1,4 +1,4 @@
-# [231. 2的幂](https://leetcode-cn.com/problems/power-of-two)
+# [231. 2 的幂](https://leetcode-cn.com/problems/power-of-two)
 
 [English Version](/solution/0200-0299/0231.Power%20of%20Two/README_EN.md)
 
@@ -25,7 +25,6 @@
 <pre><strong>输入:</strong> 218
 <strong>输出:</strong> false</pre>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
@@ -37,15 +36,52 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isPowerOfTwo(self, n: int) -> bool:
+        return n > 0 and (n & (n - 1)) == 0
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public boolean isPowerOfTwo(int n) {
+        return n > 0 && (n & (n - 1)) == 0;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isPowerOfTwo(int n) {
+        return n > 0 && (n & (n - 1)) == 0;
+    }
+};
+```
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number} n
+ * @return {boolean}
+ */
+var isPowerOfTwo = function(n) {
+    return n > 0 && (n & (n - 1)) == 0;
+};
+```
+
+### **Go**
 
+```go
+func isPowerOfTwo(n int) bool {
+    return n > 0 && (n & (n - 1)) == 0
+}
 ```
 
 ### **...**

solution/0200-0299/0231.Power of Two/README_EN.md

@@ -63,13 +63,50 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def isPowerOfTwo(self, n: int) -> bool:
+        return n > 0 and (n & (n - 1)) == 0
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public boolean isPowerOfTwo(int n) {
+        return n > 0 && (n & (n - 1)) == 0;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isPowerOfTwo(int n) {
+        return n > 0 && (n & (n - 1)) == 0;
+    }
+};
+```
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number} n
+ * @return {boolean}
+ */
+var isPowerOfTwo = function(n) {
+    return n > 0 && (n & (n - 1)) == 0;
+};
+```
+
+### **Go**
 
+```go
+func isPowerOfTwo(n int) bool {
+    return n > 0 && (n & (n - 1)) == 0
+}
 ```
 
 ### **...**

solution/0200-0299/0231.Power of Two/Solution.cpp

@@ -0,0 +1,6 @@
+class Solution {
+public:
+    bool isPowerOfTwo(int n) {
+        return n > 0 && (n & (n - 1)) == 0;
+    }
+};

solution/0200-0299/0231.Power of Two/Solution.go

@@ -0,0 +1,3 @@
+func isPowerOfTwo(n int) bool {
+    return n > 0 && (n & (n - 1)) == 0
+}

solution/0200-0299/0231.Power of Two/Solution.java

@@ -1,5 +1,5 @@
 class Solution {
     public boolean isPowerOfTwo(int n) {
-        return n > 0 && (n & -n) == n;
+        return n > 0 && (n & (n - 1)) == 0;
     }
 }

solution/0200-0299/0231.Power of Two/Solution.js

@@ -0,0 +1,7 @@
+/**
+ * @param {number} n
+ * @return {boolean}
+ */
+ var isPowerOfTwo = function(n) {
+    return n > 0 && (n & (n - 1)) == 0;
+};

solution/0200-0299/0231.Power of Two/Solution.py

@@ -1,13 +1,3 @@
 class Solution:
-    def isPowerOfTwo(self, n):
-        """
-        :type n: int
-        :rtype: bool
-        """
-        if n<=0:
-            return False
-        m=2**32
-        if m%n:
-            return False
-        else:
-            return True
+    def isPowerOfTwo(self, n: int) -> bool:
+        return n > 0 and (n & (n - 1)) == 0

solution/0300-0399/0342.Power of Four/README.md

@@ -61,15 +61,52 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isPowerOfFour(self, n: int) -> bool:
+        return n > 0 and (n & (n - 1)) == 0 and (n & 0xaaaaaaaa) == 0
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public boolean isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
+    }
+};
+```
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number} n
+ * @return {boolean}
+ */
+var isPowerOfFour = function(n) {
+    return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
+};
+```
+
+### **Go**
 
+```go
+func isPowerOfFour(n int) bool {
+    return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0
+}
 ```
 
 ### **...**

solution/0300-0399/0342.Power of Four/README_EN.md

@@ -36,13 +36,50 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def isPowerOfFour(self, n: int) -> bool:
+        return n > 0 and (n & (n - 1)) == 0 and (n & 0xaaaaaaaa) == 0
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public boolean isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
+    }
+};
+```
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number} n
+ * @return {boolean}
+ */
+var isPowerOfFour = function(n) {
+    return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
+};
+```
+
+### **Go**
 
+```go
+func isPowerOfFour(n int) bool {
+    return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0
+}
 ```
 
 ### **...**

solution/0300-0399/0342.Power of Four/Solution.cpp

@@ -0,0 +1,6 @@
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
+    }
+};