From 2ab1a21f34e34ccecea729b85b554b2039dc6ba7 Mon Sep 17 00:00:00 2001 From: haiker2011 Date: Wed, 10 Oct 2018 19:09:46 +0800 Subject: [PATCH] add #18.2 #19 python implement --- notes/面试总结.md | 132 ++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 132 insertions(+) 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