[Leetcode] 722. Remove Comments 解题报告

题目：

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 [code]/*

denotes a block comment, including line 1 and lines 6-9. The string

//

denotes line 4 as comments.
[/code]

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.
思路：

我自己采用状态转移写的代码片段1，总是无法对齐，不知道在什么时候需要换行，什么时候需要加到本行的末尾。

后来参考了网上的代码片段2，又简洁好懂，又鲁棒。

代码：

1、采用状态转移：

class Solution {
public:
vector<string> removeComments(vector<string>& source) {
bool block_comment = false;
vector<string> ans;
for (int i = 0; i < source.size(); ++i) {
if(block_comment) { // find the "*/"
size_t pos = source[i].find("*/");
if (pos != string::npos) {
source[i] = source[i].substr(pos + 2);
block_comment = false;
--i;
}
}
else {
size_t pos1 = source[i].find("//");
size_t pos2 = source[i].find("/*");
if (pos1 == string::npos && pos2 == string::npos) {
if (source[i].length() > 0) {
ans.push_back(source[i]);
}
}
else if (pos2 == string::npos || pos1 < pos2) { // treat as line comment
string s = source[i].substr(0, pos1);
if (s.length() > 0) {
ans.push_back(s);
}
}
else if (pos1 == string::npos || pos2 < pos1) { // treat as block comment
string s = source[i].substr(0, pos2);
source[i] = source[i].substr(pos2 + 2);
size_t pos3 = source[i].find("*/");
if (pos3 == string::npos) {
if (s.length() > 0) {
ans.push_back(s);
}
block_comment = true;
}
else {
source[i] = s + source[i].substr(pos3 + 2);
--i;
}
}
}
}
return ans;
}
};
2、逐字判断：

class Solution {
public:
vector<string> removeComments(vector<string>& source) {
vector<string> ans;
string s;
bool comment = false;
for(int i = 0; i < source.size(); i++) {
for(int j = 0; j < source[i].size(); j++) {
if(!comment && j + 1 < source[i].size() && source[i][j] == '/' && source[i][j+1]=='/') { // meet "//"
break;
}
else if(!comment && j + 1 < source[i].size() && source[i][j] == '/' && source[i][j+1]=='*') { // meet "/*"
comment = true;
++j;
}
else if(comment && j + 1 < source[i].size() && source[i][j] == '*' && source[i][j+1]=='/') { // meet "*/"
comment = false;
++j;
}
else if(!comment) {
s.push_back(source[i][j]);
}
}
if(!comment && s.size()) {
ans.push_back(s);
s.clear();
}
}
return ans;
}
};