Given a string representing a code snippet, you need to implement a tag validator to parse the code and return whether it is valid. A code snippet is valid if all the following rules hold:

  1. The code must be wrapped in a valid closed tag. Otherwise, the code is invalid.
  2. A closed tag (not necessarily valid) has exactly the following format : <TAG_NAME>TAG_CONTENT</TAG_NAME>. Among them, <TAG_NAME> is the start tag, and </TAG_NAME> is the end tag. The TAG_NAME in start and end tags should be the same. A closed tag is valid if and only if the TAG_NAME and TAG_CONTENT are valid.
  3. A valid TAG_NAME only contain upper-case letters, and has length in range [1,9]. Otherwise, the TAG_NAME is invalid.
  4. A valid TAG_CONTENT may contain other valid closed tags, cdata and any characters (see note1) EXCEPT unmatched <, unmatched start and end tag, and unmatched or closed tags with invalid TAG_NAME. Otherwise, the TAG_CONTENT is invalid.
  5. A start tag is unmatched if no end tag exists with the same TAG_NAME, and vice versa. However, you also need to consider the issue of unbalanced when tags are nested.
  6. < is unmatched if you cannot find a subsequent >. And when you find a < or </, all the subsequent characters until the next > should be parsed as TAG_NAME (not necessarily valid).
  7. The cdata has the following format : <![CDATA[CDATA_CONTENT]]>. The range of CDATA_CONTENT is defined as the characters between <![CDATA[ and the first subsequent ]]>.
  8. CDATA_CONTENT may contain any characters. The function of cdata is to forbid the validator to parse CDATA_CONTENT, so even it has some characters that can be parsed as tag (no matter valid or invalid), you should treat it as regular characters.

Valid Code Examples:

Input: "<DIV>This is the first line <![CDATA[<div>]]></DIV>"
Output: True
Explanation:
The code is wrapped in a closed tag : <DIV> and </DIV>.
The TAG_NAME is valid, the TAG_CONTENT consists of some characters and cdata.
Although CDATA_CONTENT has unmatched start tag with invalid TAG_NAME, it should be considered as plain text, not parsed as tag.
So TAG_CONTENT is valid, and then the code is valid. Thus return true. Input: "<DIV>>> ![cdata[]] <![CDATA[<div>]>]]>]]>>]</DIV>"
Output: True
Explanation:
We first separate the code into : start_tag|tag_content|end_tag.
start_tag -> "<DIV>"
end_tag -> "</DIV>"
tag_content could also be separated into : text1|cdata|text2.
text1 -> ">> ![cdata[]] "
cdata -> "<![CDATA[<div>]>]]>", where the CDATA_CONTENT is "<div>]>"
text2 -> "]]>>]" The reason why start_tag is NOT "<DIV>>>" is because of the rule 6.
The reason why cdata is NOT "<![CDATA[<div>]>]]>]]>" is because of the rule 7.

Invalid Code Examples:

Input: "<A>  <B> </A>   </B>"
Output: False
Explanation: Unbalanced. If "<A>" is closed, then "<B>" must be unmatched, and vice versa. Input: "<DIV> div tag is not closed <DIV>"
Output: False Input: "<DIV> unmatched < </DIV>"
Output: False Input: "<DIV> closed tags with invalid tag name <b>123</b> </DIV>"
Output: False Input: "<DIV> unmatched tags with invalid tag name </1234567890> and <CDATA[[]]> </DIV>"
Output: False Input: "<DIV> unmatched start tag <B> and unmatched end tag </C> </DIV>"
Output: False

Note:

  1. For simplicity, you could assume the input code (including the any characters mentioned above) only contain lettersdigits'<','>','/','!','[',']' and ' '.

Approach #1: Simulate. [Java]

class Solution {
public boolean isValid(String code) {
Stack<String> stack = new Stack<>();
int i, j;
for (i = 0; i < code.length(); ) {
if (i > 0 && stack.empty()) return false;
if (code.startsWith("<![CDATA[", i)) {
j = i + 9;
i = code.indexOf("]]>", j);
if (i < 0) return false;
i += 3;
} else if (code.startsWith("</", i)) {
j = i + 2;
i = code.indexOf(">", j);
if (i < 0 || i - j > 9 || i == j) return false;
for (int k = j; k < i; ++k)
if (!Character.isUpperCase(code.charAt(k))) return false;
String substr = code.substring(j, i++);
if (stack.empty() || !stack.pop().equals(substr)) return false;
} else if (code.startsWith("<", i)) {
j = i + 1;
i = code.indexOf(">", j);
if (i < 0 || i - j > 9 || i == j) return false;
for (int k = j; k < i; ++k)
if (!Character.isUpperCase(code.charAt(k))) return false;
String substr = code.substring(j, i++);
stack.push(substr);
} else {
i++;
}
}
return stack.empty();
}
}

  

Reference:

https://leetcode.com/problems/tag-validator/discuss/103368/Java-Solution%3A-Use-startsWith-and-indexOf

591. Tag Validator的更多相关文章

  1. [LeetCode] Tag Validator 标签验证器

    Given a string representing a code snippet, you need to implement a tag validator to parse the code ...

  2. [Swift]LeetCode591. 标签验证器 | Tag Validator

    Given a string representing a code snippet, you need to implement a tag validator to parse the code ...

  3. LeetCode All in One题解汇总(持续更新中...)

    突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对 ...

  4. All LeetCode Questions List 题目汇总

    All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems clas ...

  5. Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017)

    All LeetCode Questions List 题目汇总 Sorted by frequency of problems that appear in real interviews. Las ...

  6. leetcode 学习心得 (3)

    源代码地址:https://github.com/hopebo/hopelee 语言:C++ 517. Super Washing Machines You have n super washing ...

  7. leetcode hard

    # Title Solution Acceptance Difficulty Frequency     4 Median of Two Sorted Arrays       27.2% Hard ...

  8. LeetCode All in One 题目讲解汇总(转...)

    终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 如果各位看官们,大神们发现了任何错误,或是代码无法通 ...

  9. [LeetCode] Add Bold Tag in String 字符串中增添加粗标签

    Given a string s and a list of strings dict, you need to add a closed pair of bold tag <b> and ...

随机推荐

  1. idea搭spring boot项目

    Spring Boot是由Pivotal团队提供的全新框架,设计目的是用来简化新Spring应用的初始搭建以及开发过程.它主要推崇的是'消灭配置’,实现零配置. 那么,如何在idea中创建一个spri ...

  2. English-商务英文邮件例句100句

    最常用最专业的商务英文邮件例句100句——塞依SAP培训 字体大小:大 | 中 | 小2013-08-27 17:24 阅读(74) 评论(0) 分类:sap职场  1. I am writing t ...

  3. STS(eclipse)3.7.3新建项目报错:org.apache.maven.archiver.MavenArchiver.getManifest(org.apache.maven.project.MavenProject, org.apache.maven.archiver.MavenArchiveConfiguration)

    烦人的版本兼容问题 没有使用sts3.7.3系统内嵌的maven3.3.3,调整为稍高版本的maven3.5.2,懒得修改配置了. 升级eclipse插件吧. Eclipse,Help -> I ...

  4. 使用jQuery+huandlebars中with应用及with+this应用

    兼容ie8(很实用,复制过来,仅供技术参考,更详细内容请看源地址:http://www.cnblogs.com/iyangyuan/archive/2013/12/12/3471227.html) w ...

  5. 记账本,C,Github,Dao

    package dao; import java.sql.Connection; import java.sql.PreparedStatement; import java.sql.ResultSe ...

  6. Java 虚拟机概述

    虚拟机是一种抽象化的计算机,通过在实际的计算机上仿真模拟各种计算机功能来实现的.Java虚拟机有自己完善的硬体架构,如处理器.堆栈.寄存器等,还具有相应的指令系统.Java虚拟机屏蔽了与具体操作系统平 ...

  7. 利用iptables防火墙允许1521端口被连接

    今天在虚拟机上装了ora11g , 出现2个问题 1.i386的几个RPM包缺少导致安装验证不通过 2.安装完成后,本地电脑连不上虚拟机oracle , 但是能ping通 问题2: 简单的方法就是关掉 ...

  8. Curator的监听

    如果要使用类似Wather的监听功能Curator必须依赖一个jar包,Maven依赖, <dependency> <groupId>org.apache.curator< ...

  9. kubernetes 基础

    官网 kubernetes.io 有中文 中文网站  http://docs.kubernetes.org.cn kubectl 详细情况 https://kubernetes.io/docs/ref ...

  10. MAC shell ps 命令详解(转)

    ps命令为我们提供了一次性的查看进程结果,它所提供的查看结果不是动态连续的:如果想对进程时间监控,应该用top工具 Linux中的ps命令是Process Status的缩写.ps命令用来列出系统中当 ...