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@@ -79,6 +79,7 @@
 | 17   | [Letter Combinations of a Phone Number][017] | String, Backtracking             |
 | 18   | [4Sum][018]                              | Array, Hash Table, Two Pointers  |
 | 19   | [Remove Nth Node From End of List][019]  | Linked List, Two Pointers        |
+| 22   | [Generate Parentheses][022]              | String, Backtracking             |
 | 33   | [Search in Rotated Sorted Array][033]    | Arrays, Binary Search            |
 | 43   | [Multiply Strings][043]                  | Math, String                     |
 | 49   | [Group Anagrams][049]                    | Hash Table, String               |
@@ -152,6 +153,7 @@
 [017]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/017/README.md
 [018]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/018/README.md
 [019]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md
+[022]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/022/README.md
 [033]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/033/README.md
 [043]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/043/README.md
 [049]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/049/README.md
 
@@ -0,0 +1,94 @@
+# [Generate Parentheses][title]
+
+## Description
+
+Given *n* pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
+
+For example, given *n* = 3, a solution set is:
+
+```
+[
+  "((()))",
+  "(()())",
+  "(())()",
+  "()(())",
+  "()()()"
+]
+```
+
+**Tags:** String, Backtracking
+
+
+## 思路 0
+
+题意是给你 `n` 值，让你找到所有格式正确的圆括号匹配组，题目中已经给出了 `n = 3` 的所有结果。遇到这种问题，第一直觉就是用到递归或者堆栈，我们选取递归来解决，也就是 `helper` 函数的功能，从参数上来看肯定很好理解了，`leftRest` 代表还有几个左括号可以用，`rightNeed` 代表还需要几个右括号才能匹配，初始状态当然是 `rightNeed = 0, leftRest = n`，递归的终止状态就是 `rightNeed == 0 && leftRest == 0`，也就是左右括号都已匹配完毕，然后把 `str` 加入到链表中即可。
+
+```java
+class Solution {
+    public List<String> generateParenthesis(int n) {
+        List<String> list = new ArrayList<>();
+        helper(list, "", 0, n);
+        return list;
+    }
+
+    private void helper(List<String> list, String str, int rightNeed, int leftRest) {
+        if (rightNeed == 0 && leftRest == 0) {
+            list.add(str);
+            return;
+        }
+        if (rightNeed > 0) helper(list, str + ")", rightNeed - 1, leftRest);
+        if (leftRest > 0) helper(list, str + "(", rightNeed + 1, leftRest - 1);
+    }
+}
+```
+
+
+## 思路 1
+
+另一种实现方式就是迭代的思想了，我们来找寻其规律如下所示：
+
+```
+f(0): “”
+
+f(1): “(“f(0)”)”
+
+f(2): "(“f(0)”)"f(1), “(“f(1)”)”
+
+f(3): "(“f(0)”)"f(2), "(“f(1)”)"f(1), “(“f(2)”)”
+...
+```
+
+可以递推出 `f(n) = "(“f(0)”)"f(n-1) , "(“f(1)”)"f(n-2) "(“f(2)”)"f(n-3) … "(“f(i)”)“f(n-1-i) … “(f(n-1)”)”`
+
+根据如上递推式写出如下代码应该不难了吧。
+
+```java
+class Solution {
+    public List<String> generateParenthesis(int n) {
+        HashMap<Integer, List<String>> hashMap = new HashMap<>();
+        hashMap.put(0, Collections.singletonList(""));
+        for (int i = 1; i <= n; i++) {
+            List<String> list = new ArrayList<>();
+            for (int j = 0; j < i; j++) {
+                for (String fj : hashMap.get(j)) {
+                    for (String fi_j_1 : hashMap.get(i - j - 1)) {
+                        list.add("(" + fj + ")" + fi_j_1);// calculate f(i)
+                    }
+                }
+            }
+            hashMap.put(i, list);
+        }
+        return hashMap.get(n);
+    }
+}
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/generate-parentheses
+[ajl]: https://github.com/Blankj/awesome-java-leetcode
@@ -0,0 +1,53 @@
+package com.blankj.medium._022;
+
+import java.util.ArrayList;
+import java.util.Collections;
+import java.util.HashMap;
+import java.util.List;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2018/01/30
+ *     desc  :
+ * </pre>
+ */
+public class Solution {
+//    public List<String> generateParenthesis(int n) {
+//        List<String> list = new ArrayList<>();
+//        helper(list, "", 0, n);
+//        return list;
+//    }
+//
+//    private void helper(List<String> list, String str, int rightNeed, int leftRest) {
+//        if (rightNeed == 0 && leftRest == 0) {
+//            list.add(str);
+//            return;
+//        }
+//        if (rightNeed > 0) helper(list, str + ")", rightNeed - 1, leftRest);
+//        if (leftRest > 0) helper(list, str + "(", rightNeed + 1, leftRest - 1);
+//    }
+
+    public List<String> generateParenthesis(int n) {
+        HashMap<Integer, List<String>> hashMap = new HashMap<>();
+        hashMap.put(0, Collections.singletonList(""));
+        for (int i = 1; i <= n; i++) {
+            List<String> list = new ArrayList<>();
+            for (int j = 0; j < i; j++) {
+                for (String fj : hashMap.get(j)) {
+                    for (String fi_j_1 : hashMap.get(i - j - 1)) {
+                        list.add("(" + fj + ")" + fi_j_1);// calculate f(i)
+                    }
+                }
+            }
+            hashMap.put(i, list);
+        }
+        return hashMap.get(n);
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.generateParenthesis(3));
+    }
+}