feat: add solutions to lc problem: No.0693. Binary Number with Alternating Bits

Author: yanglbme

solution/0600-0699/0693.Binary Number with Alternating Bits/README.md

@@ -54,27 +54,50 @@
 	<li><code>1 <= n <= 2<sup>31</sup> - 1</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+假设 01 交替出现，那么我们可以通过错位异或将尾部全部转为 1，加 1 可以得到 2 的幂次的一个数 n（n 中只有一个位是 1），接着利用 `n & (n - 1)` 可以消除最后一位的 1。
+
+此时判断是否为 0，若是，说明假设成立，是 01 交替串。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def hasAlternatingBits(self, n: int) -> bool:
+        n = (n ^ (n >> 1)) + 1
+        return (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 hasAlternatingBits(int n) {
+        n = (n ^ (n >> 1)) + 1;
+        return (n & (n - 1)) == 0;
+    }
+}
+```
+
+### **C++**
 
+```cpp
+class Solution {
+public:
+    bool hasAlternatingBits(int n) {
+        n ^= (n >> 1);
+        return (n & ((long) n + 1)) == 0;
+    }
+};
 ```
 
 ### **...**

solution/0600-0699/0693.Binary Number with Alternating Bits/README_EN.md

@@ -58,13 +58,33 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def hasAlternatingBits(self, n: int) -> bool:
+        n = (n ^ (n >> 1)) + 1
+        return (n & (n - 1)) == 0
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public boolean hasAlternatingBits(int n) {
+        n = (n ^ (n >> 1)) + 1;
+        return (n & (n - 1)) == 0;
+    }
+}
+```
+
+### **C++**
 
+```cpp
+class Solution {
+public:
+    bool hasAlternatingBits(int n) {
+        n ^= (n >> 1);
+        return (n & ((long) n + 1)) == 0;
+    }
+};
 ```
 
 ### **...**

solution/0600-0699/0693.Binary Number with Alternating Bits/Solution.cpp

@@ -0,0 +1,7 @@
+class Solution {
+public:
+    bool hasAlternatingBits(int n) {
+        n ^= (n >> 1);
+        return (n & ((long) n + 1)) == 0;
+    }
+};

solution/0600-0699/0693.Binary Number with Alternating Bits/Solution.java

@@ -0,0 +1,6 @@
+class Solution {
+    public boolean hasAlternatingBits(int n) {
+        n = (n ^ (n >> 1)) + 1;
+        return (n & (n - 1)) == 0;
+    }
+}

solution/0600-0699/0693.Binary Number with Alternating Bits/Solution.py

@@ -0,0 +1,4 @@
+class Solution:
+    def hasAlternatingBits(self, n: int) -> bool:
+        n = (n ^ (n >> 1)) + 1
+        return (n & (n - 1)) == 0