feat: add solutions to lc problem: No.0201

No.0201.Bitwise AND of Numbers Range

Author: yanglbme

solution/0200-0299/0201.Bitwise AND of Numbers Range/README.md

@@ -43,22 +43,74 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：位运算**
+
+题目可以转换为求数字的公共二进制前缀。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def rangeBitwiseAnd(self, left: int, right: int) -> int:
+        while left < right:
+            right &= (right - 1)
+        return right
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int rangeBitwiseAnd(int left, int right) {
+        while (left < right) {
+            right &= (right - 1);
+        }
+        return right;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int rangeBitwiseAnd(int left, int right) {
+        while (left < right) right &= (right - 1);
+        return right;
+    }
+};
+```
+
+### **Go**
+
+```go
+func rangeBitwiseAnd(left int, right int) int {
+	for left < right {
+		right &= (right - 1)
+	}
+	return right
+}
+```
 
+### **C#**
+
+```cs
+public class Solution {
+    public int RangeBitwiseAnd(int left, int right) {
+        while (left < right)
+        {
+            right &= (right - 1);
+        }
+        return right;
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0201.Bitwise AND of Numbers Range/README_EN.md

@@ -42,13 +42,61 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def rangeBitwiseAnd(self, left: int, right: int) -> int:
+        while left < right:
+            right &= (right - 1)
+        return right
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int rangeBitwiseAnd(int left, int right) {
+        while (left < right) {
+            right &= (right - 1);
+        }
+        return right;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int rangeBitwiseAnd(int left, int right) {
+        while (left < right) right &= (right - 1);
+        return right;
+    }
+};
+```
+
+### **Go**
+
+```go
+func rangeBitwiseAnd(left int, right int) int {
+	for left < right {
+		right &= (right - 1)
+	}
+	return right
+}
+```
 
+### **C#**
+
+```cs
+public class Solution {
+    public int RangeBitwiseAnd(int left, int right) {
+        while (left < right)
+        {
+            right &= (right - 1);
+        }
+        return right;
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0201.Bitwise AND of Numbers Range/Solution.cpp

@@ -0,0 +1,7 @@
+class Solution {
+public:
+    int rangeBitwiseAnd(int left, int right) {
+        while (left < right) right &= (right - 1);
+        return right;
+    }
+};

solution/0200-0299/0201.Bitwise AND of Numbers Range/Solution.cs

@@ -1,13 +1,9 @@
 public class Solution {
-    public int RangeBitwiseAnd(int m, int n) {
-        var and = m & n;
-        var xor = m ^ n;
-        var temp = 0;
-        while (xor > 0)
+    public int RangeBitwiseAnd(int left, int right) {
+        while (left < right)
         {
-            temp = (temp << 1) + 1;
-            xor >>= 1;
+            right &= (right - 1);
         }
-        return and & (~temp);
+        return right;
     }
 }

solution/0200-0299/0201.Bitwise AND of Numbers Range/Solution.go

@@ -0,0 +1,6 @@
+func rangeBitwiseAnd(left int, right int) int {
+	for left < right {
+		right &= (right - 1)
+	}
+	return right
+}

solution/0200-0299/0201.Bitwise AND of Numbers Range/Solution.java

@@ -1,8 +1,8 @@
 class Solution {
-    public int rangeBitwiseAnd(int m, int n) {
-        while (m < n) {
-            n &= n - 1;
+    public int rangeBitwiseAnd(int left, int right) {
+        while (left < right) {
+            right &= (right - 1);
         }
-        return n;
+        return right;
     }
-}
+}

solution/0200-0299/0201.Bitwise AND of Numbers Range/Solution.py

@@ -0,0 +1,5 @@
+class Solution:
+    def rangeBitwiseAnd(self, left: int, right: int) -> int:
+        while left < right:
+            right &= (right - 1)
+        return right

solution/0600-0699/0699.Falling Squares/README.md

@@ -90,6 +90,13 @@ _aa___a
 -   线段树的每个叶子节点代表一个长度为 1 的元区间 `[x, x]`；
 -   对于每个内部节点 `[l, r]`，它的左儿子是 `[l, mid]`，右儿子是 `[mid + 1, r]`, 其中 `mid = ⌊(l + r) / 2⌋` (即向下取整)。
 
+对于本题，线段树节点维护的信息有：
+
+1. 区间中方块的最大高度 v
+1. 懒标记 add
+
+另外，由于数轴范围很大，达到 10^8，因此我们采用动态开点。
+
 <!-- tabs:start -->
 
 ### **Python3**