diff --git a/notes/面试总结.md b/notes/面试总结.md
index a8a47dd3..82974d9a 100644
--- a/notes/面试总结.md
+++ b/notes/面试总结.md
@@ -20,6 +20,8 @@
* [16. 数值的整数次方](#16-数值的整数次方)
* [17. 打印从 1 到最大的 n 位数](#17-打印从-1-到最大的-n-位数)
* [18.1 在 O(1) 时间内删除链表节点](#181-在-o1-时间内删除链表节点)
+* [18.2 删除链表中重复的结点](#182-删除链表中重复的结点)
+* [19. 正则表达式匹配](#19-正则表达式匹配)
* [参考文献](#参考文献)
@@ -1439,6 +1441,136 @@ if __name__ == '__main__':
else:
print ('wrong')
```
+# 18.2 删除链表中重复的结点
+
+[NowCoder](https://www.nowcoder.com/practice/fc533c45b73a41b0b44ccba763f866ef?tpId=13&tqId=11209&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+
+
+## 解题描述
+
+```java
+public ListNode deleteDuplication(ListNode pHead) {
+ if (pHead == null || pHead.next == null)
+ return pHead;
+ ListNode next = pHead.next;
+ if (pHead.val == next.val) {
+ while (next != null && pHead.val == next.val)
+ next = next.next;
+ return deleteDuplication(next);
+ } else {
+ pHead.next = deleteDuplication(pHead.next);
+ return pHead;
+ }
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class ListNode:
+# def __init__(self, x):
+# self.val = x
+# self.next = None
+class Solution:
+ def deleteDuplication(self, pHead):
+ # write code here
+ if not pHead or not pHead.next:
+ return pHead
+ next = pHead.next
+ if pHead.val == next.val:
+ while next and pHead.val == next.val:
+ next = next.next
+ return self.deleteDuplication(next)
+ else:
+ pHead.next = self.deleteDuplication(pHead.next)
+ return pHead
+```
+
+# 19. 正则表达式匹配
+
+[NowCoder](https://www.nowcoder.com/practice/45327ae22b7b413ea21df13ee7d6429c?tpId=13&tqId=11205&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+请实现一个函数用来匹配包括 '.' 和 '\*' 的正则表达式。模式中的字符 '.' 表示任意一个字符，而 '\*' 表示它前面的字符可以出现任意次（包含 0 次）。
+
+在本题中，匹配是指字符串的所有字符匹配整个模式。例如，字符串 "aaa" 与模式 "a.a" 和 "ab\*ac\*a" 匹配，但是与 "aa.a" 和 "ab\*a" 均不匹配。
+
+## 解题思路
+
+应该注意到，'.' 是用来当做一个任意字符，而 '\*' 是用来重复前面的字符。这两个的作用不同，不能把 '.' 的作用和 '\*' 进行类比，从而把它当成重复前面字符一次。
+
+```java
+public boolean match(char[] str, char[] pattern) {
+
+ int m = str.length, n = pattern.length;
+ boolean[][] dp = new boolean[m + 1][n + 1];
+
+ dp[0][0] = true;
+ for (int i = 1; i <= n; i++)
+ if (pattern[i - 1] == '*')
+ dp[0][i] = dp[0][i - 2];
+
+ for (int i = 1; i <= m; i++)
+ for (int j = 1; j <= n; j++)
+ if (str[i - 1] == pattern[j - 1] || pattern[j - 1] == '.')
+ dp[i][j] = dp[i - 1][j - 1];
+ else if (pattern[j - 1] == '*')
+ if (pattern[j - 2] == str[i - 1] || pattern[j - 2] == '.') {
+ dp[i][j] |= dp[i][j - 1]; // a* counts as single a
+ dp[i][j] |= dp[i - 1][j]; // a* counts as multiple a
+ dp[i][j] |= dp[i][j - 2]; // a* counts as empty
+ } else
+ dp[i][j] = dp[i][j - 2]; // a* only counts as empty
+
+ return dp[m][n];
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+class Solution:
+ # s, pattern都是字符串
+ def match(self, s, pattern):
+ # write code here
+ if (len(s) == 0 and len(pattern) == 0):
+ return True
+ if (len(s) > 0 and len(pattern) == 0):
+ return False
+ if (len(pattern) > 1 and pattern[1] == '*'):
+ if (len(s) > 0 and (s[0] == pattern[0] or pattern[0] == '.')):
+ return (self.match(s, pattern[2:]) or self.match(s[1:], pattern[2:]) or self.match(s[1:], pattern))
+ else:
+ return self.match(s, pattern[2:])
+ if (len(s) > 0 and (pattern[0] == '.' or pattern[0] == s[0])):
+ return self.match(s[1:], pattern[1:])
+ return False
+```
+
+```python
+# -*- coding:utf-8 -*-
+import re
+class Solution:
+ # s, pattern都是字符串
+ def match(self, s, pattern):
+ # write code here
+ if s == None or pattern == None:
+ return False
+ if s == "" and pattern == "":
+ return True
+ if pattern == "" and s != "":
+ return False
+
+ res = re.compile(str(pattern)+"$")
+ result = res.match(str(s))
+
+ if result is None:
+ return False
+ else:
+ return True
+```
# 参考文献
