Google中国2014校园招聘笔试Round B China New Grad Test Problem E. Ignore all my comments
Problem
Good programmers write fabulous comments. Igor is a programmer and he likes the old C-style comments in /* ... */ blocks. For him, it would be ideal if he could use this style as a uniform comment format for all programming languages or even documents, for example Python, Haskell or HTML/XML documents.
Making this happen doesn't seem too difficult to Igor. What he will need is a comment pre-processor that removes all the comment blocks in /*, followed by comment text, and by another */. Then the processed text can be handed over to the compiler/document renderer to which it belongs—whatever it is.
Igor's pre-processor isn't quite that simple, though. Here are some cool things it does:
printf("Hello /* a comment /* a comment inside comment */
inside /* another comment inside comment */
string */ world");
After the pre-process step, it becomes: printf("Hello world");
"/*...*/", a constant number 12/*...*/34 or even in a character escape \/*...*/n Or more formally:
text: text-piece text-piece remaining-text text-piece: char-sequence-without-/* empty-string remaining-text: comment-block text comment-block: /* comment-content */ comment-content: comment-piece comment-piece remaining-comment comment-piece: char-sequence-without-/*-or-*/ empty-string remaining-comment: comment-block comment-content char: letters digits punctuations whitespaces
Our pre-processor, given a text, removes all comment-block instances as specified.
Notes
//*no recursion*/* file header */should generate:
/* file header */
* character in any /* or /*cannot be re-used in another /* or */. For example the following does NOT form a proper comment block /*/
Input
A text document with comment blocks in /* and */. The input file is valid. It follows the specification of text in the problem statement. The input file always terminates with a newline symbol.
Output
We only have one test case for this problem. First we need to output the following line.
Case #1:Then, print the document with all comments removed, in the way specified in the problem statements. Don't remove any spaces or empty lines outside comments.
Limits
The input program contains only:
a-z, A-Z,0-9~ ! @ # % ^ & * ( ) - + = : ; " ' < > , . ? | / \ { } [ ] _Small dataset
The small input contains a program of less than 2k bytes.
Large dataset
The large input contains a program of less than 100k bytes.
Sample
| Input |
|
//*no recursion*/* file header
***********/************
* Sample input program *
**********/*************
*/
int spawn_workers(int worker_count) {
/* The block below is supposed to spawn 100 workers.
But it creates many more.
Commented until I figure out why.
for (int i = 0; i < worker_count; ++i) {
if(!fork()) {
/* This is the worker. Start working. */
do_work();
}
}
*/
return 0; /* successfully spawned 100 workers */
}
int main() {
printf("Hello /*a comment inside string*/ world");
int worker_count = 0/*octal number*/144;
if (spawn_workers(worker_count) != 0) {
exit(-1);
}
return 0;
}
|
|
| |
|
| Output |
|
Case #1:
/* file header
************************
*/
int spawn_workers(int worker_count) {
return 0;
}
int main() {
printf("Hello world");
int worker_count = 0144;
if (spawn_workers(worker_count) != 0) {
exit(-1);
}
return 0;
}
|
类型：字符串 难度：2
题意：给一段文本（多行），文本中包含/* ... */的注释段，要求把注释内容去掉，输出结果。但是这个注释和一般的略有不同，有两个特点：
（1）能够嵌套，类似括号的用法
（2）可能出现在文本任何地方，如"/*...*/"，12/*...*/34
还有两点要注意的是：
（1）去掉注释文本后形成的/*或*/不算注释，如
//*no recursion*/* file header */
去掉注释后为：
/* file header */
（2）注释中的*只能被一个/*或*/所用，如/*/不是合法的注释
分析：题目本身看似难度不大，但是通过率很低，归根结底还是细节问题较多（其实总体看google的笔试，都不是需要很难的算法才能解出的，考察的都是细节的把握）
基本思路：
（1）遍历文本，用一个栈nest存储/*的位置，遇到/*则入栈，遇到*/则将nest顶端的/*出栈
（2）用另一个栈del记录互不包含重叠的最外层/*..*/对的位置，即当nest出栈后，栈为空，证明最后出栈的是最外层的/*...*/，将它们的位置入del栈
（3）遍历结束后，del内位置对依次出栈，删除对应范围的文本（先删后边的保证前面的位置下标不变，这也是用栈del的原因）
两个细节：
（1）遍历时先找'/'，再找前一个是不是'*'，即找*/，因为要先和前面的/*匹配。再找后一个是不是'*'
（2）注意标记用过的'*'，防止/*/情况中'*'被前一个'/'用完又被后一个'/'用。
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
using namespace std;
typedef struct
{
int l,r;
}range;
int main()
{
freopen("E-large-practice.in","r",stdin);
freopen("E-large-practice.out","w",stdout);
string tmp,text;
while(getline(cin,tmp,'\n'))
{
text+=tmp;
text+="\n";
}
vector<int> nest;
vector<range> del;
nest.clear();
del.clear();
int pos = -1;
for(int i=0; i<text.size(); i++)
{
if(text[i]=='/')
{
if(i>0 && text[i-1]=='*' && !nest.empty() && pos!=(i-1))
{
int left = nest.back();
nest.pop_back();
if(nest.empty())
{
range tn;
tn.l = left;
tn.r = i;
del.push_back(tn);
}
}
else if(i<text.size()-1 && text[i+1]=='*')
{
nest.push_back(i);
pos = i+1;
}
}
}
while(!del.empty())
{
range td = del.back();
text.erase(td.l,td.r-td.l+1);
del.pop_back();
}
cout<<"Case #1:\n"<<text;
}