update

Author: jasoncao11

10. 合并两个有序链表.md

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+**将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。**
+
+```
+输入：l1 = [1,2,4], l2 = [1,3,4]
+输出：[1,1,2,3,4,4]
+```
+
+```
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+        # 终止条件，直到两个链表都空
+        if not list1:
+            return list2
+        if not list2:
+            return list1
+        # 递归调用
+        if list1.val < list2.val:
+            list1.next = self.mergeTwoLists(list1.next, list2)
+            return list1
+        else:
+            list2.next = self.mergeTwoLists(list1, list2.next)
+            return list2
+```

11. 括号生成.md

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+**数字 n 代表生成括号的对数，请你设计一个函数，用于能够生成所有可能的并且有效的括号组合。**
+
+```
+输入：n = 3
+输出：["((()))","(()())","(())()","()(())","()()()"]
+```
+
+Ref: https://leetcode.cn/problems/generate-parentheses/solution/hui-su-suan-fa-by-liweiwei1419/
+
+![algo5](./images/algo5.jpg)
+
+```
+class Solution:
+    def generateParenthesis(self, n: int) -> List[str]:
+        #回溯法
+        res = []
+        #当前括号组合
+        cur = ''
+        #当前剩余左右括号数
+        left = right = n
+        #构造递归
+        def dfs(left, right, cur):
+            #递归终止条件
+            if left==0 and right==0:
+                res.append(cur)
+                return
+            #情况1
+            if left>0:
+                dfs(left-1, right, cur+'(')
+            #情况1
+            if right > left:
+                dfs(left, right-1, cur+')')
+        #调用递归
+        dfs(left, right, cur)
+        #返回要优化的目标
+        return res
+```