add #18.2 #19 python implement

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haiker2011 2018-10-10 19:09:46 +08:00
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commit 2ab1a21f34

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@ -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)
<!-- GFM-TOC -->
@ -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
<div align="center"> <img src="../pics//8433fbb2-c35c-45ef-831d-e3ca42aebd51.png" width="500"/> </div><br>
## 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