feat: add solutions to lc problem: No.0693

No.0693.Binary Number with Alternating Bits

Author: yanglbme

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

@@ -44,7 +44,13 @@
 
 <!-- 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。
+**方法一：模拟**
+
+n 循环右移直至为 0，依次检测 n 的二进制位是否交替出现。若循环过程中发现 0、1 没有交替出现，直接返回 false。否则循环结束返回 true。
+
+**方法二：位运算**
+
+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。
 
@@ -57,8 +63,21 @@
 ```python
 class Solution:
     def hasAlternatingBits(self, n: int) -> bool:
-        n = (n ^ (n >> 1)) + 1
-        return (n & (n - 1)) == 0
+        prev = -1
+        while n:
+            curr = n & 1
+            if prev == curr:
+                return False
+            prev = curr
+            n >>= 1
+        return True
+```
+
+```python
+class Solution:
+    def hasAlternatingBits(self, n: int) -> bool:
+        n ^= (n >> 1)
+        return (n & (n + 1)) == 0
 ```
 
 ### **Java**
@@ -68,14 +87,48 @@ class Solution:
 ```java
 class Solution {
     public boolean hasAlternatingBits(int n) {
-        n = (n ^ (n >> 1)) + 1;
-        return (n & (n - 1)) == 0;
+        int prev = -1;
+        while (n != 0) {
+            int curr = n & 1;
+            if (prev == curr) {
+                return false;
+            }
+            prev = curr;
+            n >>= 1;
+        }
+        return true;
+    }
+}
+```
+
+```java
+class Solution {
+    public boolean hasAlternatingBits(int n) {
+        n ^= (n >> 1);
+        return (n & (n + 1)) == 0;
     }
 }
 ```
 
 ### **C++**
 
+```cpp
+class Solution {
+public:
+    bool hasAlternatingBits(int n) {
+        int prev = -1;
+        while (n)
+        {
+            int curr = n & 1;
+            if (prev == curr) return false;
+            prev = curr;
+            n >>= 1;
+        }
+        return true;
+    }
+};
+```
+
 ```cpp
 class Solution {
 public:
@@ -115,6 +168,30 @@ impl Solution {
 }
 ```
 
+### **Go**
+
+```go
+func hasAlternatingBits(n int) bool {
+	prev := -1
+	for n != 0 {
+		curr := n & 1
+		if prev == curr {
+			return false
+		}
+		prev = curr
+		n >>= 1
+	}
+	return true
+}
+```
+
+```go
+func hasAlternatingBits(n int) bool {
+	n ^= (n >> 1)
+	return (n & (n + 1)) == 0
+}
+```
+
 ### **...**
 
 ```

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

@@ -45,23 +45,70 @@
 ```python
 class Solution:
     def hasAlternatingBits(self, n: int) -> bool:
-        n = (n ^ (n >> 1)) + 1
-        return (n & (n - 1)) == 0
+        prev = -1
+        while n:
+            curr = n & 1
+            if prev == curr:
+                return False
+            prev = curr
+            n >>= 1
+        return True
+```
+
+```python
+class Solution:
+    def hasAlternatingBits(self, n: int) -> bool:
+        n ^= (n >> 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;
+        int prev = -1;
+        while (n != 0) {
+            int curr = n & 1;
+            if (prev == curr) {
+                return false;
+            }
+            prev = curr;
+            n >>= 1;
+        }
+        return true;
+    }
+}
+```
+
+```java
+class Solution {
+    public boolean hasAlternatingBits(int n) {
+        n ^= (n >> 1);
+        return (n & (n + 1)) == 0;
     }
 }
 ```
 
 ### **C++**
 
+```cpp
+class Solution {
+public:
+    bool hasAlternatingBits(int n) {
+        int prev = -1;
+        while (n)
+        {
+            int curr = n & 1;
+            if (prev == curr) return false;
+            prev = curr;
+            n >>= 1;
+        }
+        return true;
+    }
+};
+```
+
 ```cpp
 class Solution {
 public:
@@ -101,6 +148,30 @@ impl Solution {
 }
 ```
 
+### **Go**
+
+```go
+func hasAlternatingBits(n int) bool {
+	prev := -1
+	for n != 0 {
+		curr := n & 1
+		if prev == curr {
+			return false
+		}
+		prev = curr
+		n >>= 1
+	}
+	return true
+}
+```
+
+```go
+func hasAlternatingBits(n int) bool {
+	n ^= (n >> 1)
+	return (n & (n + 1)) == 0
+}
+```
+
 ### **...**
 
 ```

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

@@ -0,0 +1,4 @@
+func hasAlternatingBits(n int) bool {
+	n ^= (n >> 1)
+	return (n & (n + 1)) == 0
+}

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

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

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

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