feat: add python and java solutions to lcci question: 08.09.Bracket

添加《程序员面试金典》题解：面试题 08.09. 括号

Author: yanglbme

lcci/08.09.Bracket/README.md

@@ -25,16 +25,62 @@
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
+递归求解。其中，`left` 表示剩余的 `(`，`right` 表示剩余的 `)`。
+
+- 当 `left` > `right` 时，说明 state 中 `(` 少于 `)`，不是合法组合，直接剪枝；
+- 当 `right` == 0 时，说明 state 组合完毕；
+- 当 `left` > 0 时，此时可往 state 添加一个 `(`；
+- 当 `right` > 0 时，此时可往 state 添加一个 `)`。
 
-```python
 
+```python
+class Solution:
+    def generateParenthesis(self, n: int) -> List[str]:
+        res = []
+        def generate(state, left, right):
+            # 剩余的`(`多于`)`
+            if left > right:
+                return
+            if right == 0:
+                res.append(state)
+                return
+            if left > 0:
+                generate(state + '(', left - 1, right)
+            if right > 0:
+                generate(state + ')', left, right - 1)
+        generate('', n, n)
+        return res
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    List<String> res;
+
+    public List<String> generateParenthesis(int n) {
+        res = new ArrayList<>();
+        generate("", n, n);
+        return res;
+    }
 
+    private void generate(String state, int left, int right) {
+        if (left > right) {
+            return;
+        }
+        if (right == 0) {
+            res.add(state);
+            return;
+        }
+        if (left > 0) {
+            generate(state + "(", left - 1, right);
+        }
+        if (right > 0) {
+            generate(state + ")", left, right - 1);
+        }
+    }
+}
 ```
 
 ### ...

lcci/08.09.Bracket/README_EN.md

@@ -1,40 +1,93 @@
-# [08.09. Bracket](https://leetcode-cn.com/problems/bracket-lcci)
-
-## Description
-<p>Implement an algorithm to print all valid (e.g., properly opened and closed) combinations of n pairs of parentheses.</p>
-
-<p>Note: The result set should not contain duplicated subsets.</p>
-
-<p>For example, given&nbsp;n = 3, the result should be:</p>
-
-<pre>
-[
-  &quot;((()))&quot;,
-  &quot;(()())&quot;,
-  &quot;(())()&quot;,
-  &quot;()(())&quot;,
-  &quot;()()()&quot;
-]
-</pre>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [08.09. Bracket](https://leetcode-cn.com/problems/bracket-lcci)
+
+## Description
+<p>Implement an algorithm to print all valid (e.g., properly opened and closed) combinations of n pairs of parentheses.</p>
+
+
+
+<p>Note: The result set should not contain duplicated subsets.</p>
+
+
+
+<p>For example, given&nbsp;n = 3, the result should be:</p>
+
+
+
+<pre>
+
+[
+
+  &quot;((()))&quot;,
+
+  &quot;(()())&quot;,
+
+  &quot;(())()&quot;,
+
+  &quot;()(())&quot;,
+
+  &quot;()()()&quot;
+
+]
+
+</pre>
+
+
+
+
+## Solutions
+
+
+### Python3
+
+```python
+class Solution:
+    def generateParenthesis(self, n: int) -> List[str]:
+        res = []
+        def generate(state, left, right):
+            if left > right:
+                return
+            if right == 0:
+                res.append(state)
+                return
+            if left > 0:
+                generate(state + '(', left - 1, right)
+            if right > 0:
+                generate(state + ')', left, right - 1)
+        generate('', n, n)
+        return res
+```
+
+### Java
+
+```java
+class Solution {
+    List<String> res;
+
+    public List<String> generateParenthesis(int n) {
+        res = new ArrayList<>();
+        generate("", n, n);
+        return res;
+    }
+
+    private void generate(String state, int left, int right) {
+        if (left > right) {
+            return;
+        }
+        if (right == 0) {
+            res.add(state);
+            return;
+        }
+        if (left > 0) {
+            generate(state + "(", left - 1, right);
+        }
+        if (right > 0) {
+            generate(state + ")", left, right - 1);
+        }
+    }
+}
+```
+
+### ...
+```
+
+```

lcci/08.09.Bracket/Solution.java

@@ -0,0 +1,25 @@
+class Solution {
+    List<String> res;
+
+    public List<String> generateParenthesis(int n) {
+        res = new ArrayList<>();
+        generate("", n, n);
+        return res;
+    }
+
+    private void generate(String state, int left, int right) {
+        if (left > right) {
+            return;
+        }
+        if (right == 0) {
+            res.add(state);
+            return;
+        }
+        if (left > 0) {
+            generate(state + "(", left - 1, right);
+        }
+        if (right > 0) {
+            generate(state + ")", left, right - 1);
+        }
+    }
+}

lcci/08.09.Bracket/Solution.py

@@ -0,0 +1,15 @@
+class Solution:
+    def generateParenthesis(self, n: int) -> List[str]:
+        res = []
+        def generate(state, left, right):
+            if left > right:
+                return
+            if right == 0:
+                res.append(state)
+                return
+            if left > 0:
+                generate(state + '(', left - 1, right)
+            if right > 0:
+                generate(state + ')', left, right - 1)
+        generate('', n, n)
+        return res