题目如下:

Given a C++ program, remove comments from it. The program source is an array where source[i] is the i-th line of the source code. This represents the result of splitting the original source code string by the newline character \n.

In C++, there are two types of comments, line comments, and block comments.

The string // denotes a line comment, which represents that it and rest of the characters to the right of it in the same line should be ignored.

The string /* denotes a block comment, which represents that all characters until the next (non-overlapping) occurrence of */ should be ignored. (Here, occurrences happen in reading order: line by line from left to right.) To be clear, the string /*/does not yet end the block comment, as the ending would be overlapping the beginning.

The first effective comment takes precedence over others: if the string // occurs in a block comment, it is ignored. Similarly, if the string /* occurs in a line or block comment, it is also ignored.

If a certain line of code is empty after removing comments, you must not output that line: each string in the answer list will be non-empty.

There will be no control characters, single quote, or double quote characters. For example, source = "string s = "/* Not a comment. */";" will not be a test case. (Also, nothing else such as defines or macros will interfere with the comments.)

It is guaranteed that every open block comment will eventually be closed, so /* outside of a line or block comment always starts a new comment.

Finally, implicit newline characters can be deleted by block comments. Please see the examples below for details.

After removing the comments from the source code, return the source code in the same format.

Example 1:

Input:
source = ["/*Test program */", "int main()", "{ ", " // variable declaration ", "int a, b, c;", "/* This is a test", " multiline ", " comment for ", " testing */", "a = b + c;", "}"] The line by line code is visualized as below:
/*Test program */
int main()
{
// variable declaration
int a, b, c;
/* This is a test
multiline
comment for
testing */
a = b + c;
} Output: ["int main()","{ "," ","int a, b, c;","a = b + c;","}"] The line by line code is visualized as below:
int main()
{ int a, b, c;
a = b + c;
} Explanation:
The string /* denotes a block comment, including line 1 and lines 6-9. The string // denotes line 4 as comments.

Example 2:

Input:
source = ["a/*comment", "line", "more_comment*/b"]
Output: ["ab"]
Explanation: The original source string is "a/*comment\nline\nmore_comment*/b", where we have bolded the newline characters.
After deletion, the implicit newline characters are deleted, leaving the string "ab",
which when delimited by newline characters becomes ["ab"].

Note:

  • The length of source is in the range [1, 100].
  • The length of source[i] is in the range [0, 80].
  • Every open block comment is eventually closed.
  • There are no single-quote, double-quote, or control characters in the source code.

解题思路:这种题目还是很烦的,要考虑的情况比较多。我的方法相对简单粗暴一点,首先设置一个定界符,例如:'#$%@'。然后把所有的换行符都替换成定界符,这样相当于把代码都合并到一行。接下来查找下标最小的'//'和'/*',如果'/*'的下标更小找出在后面最近的'*/',删除掉之间所有字符;否则,找出'//'后面最近的定界符,并且删除到之间的所有字符。循环操作直到所有合法的//'和'/*'都删除掉为止,最后把定界符再替换回换行符。

代码如下:

class Solution(object):
def removeComments(self, source):
"""
:type source: List[str]
:rtype: List[str]
"""
newLine = ''
delimiter = '#$%@'
for i in source:
newLine += i
newLine += delimiter while True:
linecommentStart = newLine.find('//')
blockCommentStart = newLine.find('/*')
if linecommentStart == -1 and blockCommentStart == -1:
break
elif linecommentStart == -1:
blockCommentEnd = newLine.find('*/',blockCommentStart+2)
if blockCommentEnd == -1:
break
newLine = newLine[:blockCommentStart] + newLine[blockCommentEnd+2:]
elif blockCommentStart == -1:
linecommentEnd = newLine.find(delimiter,linecommentStart)
newLine = newLine[:linecommentStart] + newLine[linecommentEnd + 4:]
else:
if linecommentStart < blockCommentStart:
linecommentEnd = newLine.find(delimiter, linecommentStart)
newLine = newLine[:linecommentStart] + newLine[linecommentEnd:]
else:
blockCommentEnd = newLine.find('*/', blockCommentStart + 2)
if blockCommentEnd != -1:
newLine = newLine[:blockCommentStart] + newLine[blockCommentEnd + 2:]
else:
linecommentEnd = newLine.find(delimiter, linecommentStart)
newLine = newLine[:linecommentStart] + newLine[linecommentEnd:]
def filterEmpty(n):
return len(n) > 0 return list(filter(filterEmpty, newLine.split(delimiter)))

【leetcode】722. Remove Comments的更多相关文章

  1. 【LeetCode】722. Remove Comments 解题报告(Python)

    [LeetCode]722. Remove Comments 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/remove-c ...

  2. 【LeetCode】402. Remove K Digits 解题报告(Python)

    [LeetCode]402. Remove K Digits 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http: ...

  3. 【leetcode】 26. Remove Duplicates from Sorted Array

    @requires_authorization @author johnsondu @create_time 2015.7.22 18:58 @url [remove dublicates from ...

  4. 【leetcode】1233. Remove Sub-Folders from the Filesystem

    题目如下: Given a list of folders, remove all sub-folders in those folders and return in any order the f ...

  5. 【LeetCode】26. Remove Duplicates from Sorted Array 解题报告(Python&C++&Java)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 双指针 日期 [LeetCode] https:// ...

  6. 【LeetCode】27. Remove Element 解题报告(Python & Java)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 双指针 记录起始位置 日期 题目地址:https:/ ...

  7. 【LeetCode】1047. Remove All Adjacent Duplicates In String 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 栈 日期 题目地址:https://leetcode ...

  8. 【LeetCode】316. Remove Duplicate Letters 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...

  9. 【LeetCode】203. Remove Linked List Elements 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 双指针 递归 日期 题目地址:https://lee ...

随机推荐

  1. DB-MD:MD/主数据

    ylbtech-DB-MD:MD/主数据 主数据(MD Master Data)指系统间共享数据(例如,客户.供应商.账户和组织部门相关数据).与记录业务活动,波动较大的交易数据相比,主数据(也称基准 ...

  2. Oracle--索引视图序列等对象

    ---恢复内容开始--- 索引 与表类似,不仅需要在DD中保存索引的定义,还需要在表空间为它分配实际的存储空间. 将索引和对应的表分别存放在不同硬盘的不同表空间中能够提高查询的速度,因为Oracle能 ...

  3. Linux_自制系统服务启动脚本

    目录 目录 前言 Case语句 Apache 启动脚本 Postfix service 启停脚本 前言 在Linux的某些系统服务中,需要自己定制启动服务的脚本.通常会使用Cash语句来实现. Cas ...

  4. Oracle 一条sql插入多条数据

    Oracle一次插入多条数据. 表结构: create table aa ( ID NUMBER(11) PRIMARY KEY, NAME VARCHAR2(20) ) 第一种方式: insert ...

  5. JSP 学习笔记1

    JSP 学习笔记   JSP是Servlet的一种特殊形式,每个JSP页面就是一个Servlet实例--JSP页面有系统编译成Servlet,Servlet再负责响应用户请求. 1.JSP注释 < ...

  6. 【FICO系列】SAP FICO模块-固定资产月结的注意点

    公众号:SAP Technical 本文作者:matinal 原文出处:http://www.cnblogs.com/SAPmatinal/ 原文链接:[FICO系列]SAP FICO模块-固定资产月 ...

  7. Mac005--VS&webstorm前端开发工具安装

    Mac--Visual studio Code工具安装(企业常用) 安装网址:https://code.visualstudio.com/download 设置格式: 1.配置工作区与终端字体大小 常 ...

  8. Netty内存池及命中缓存的分配

    内存池的内存规格: 在前面的源码分析过程中,关于内存规格大小我们应该还有些印象.其实在Netty 内存池中主要设置了四种规格大小的内存:tiny 是指0-512Byte 之间的规格大小,small 是 ...

  9. css画百分比圆环

    html: <div class="circle"> <div class="percent-circle percent-circle-left&qu ...

  10. [POJ3612] Telephone Wire(暴力dp+剪枝)

    [POJ3612] Telephone Wire(暴力dp+剪枝) 题面 有N根电线杆,初始高度为h[i],要给相邻的两根连线.可以选择拔高其中一部分电线杆,把一根电线杆拔高\(\Delta H\)的 ...