[LeetCode] Remove Comments 移除注释

 

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.

 

这道题让咱们移除代码中的注释部分，就是写代码中常常遇到的两种注释，单行注释和多行注释，也能够叫块注释，固然最最重要的就是要找到这两种注释的起始标识符"//"和"/*"，注意它们二者之间存在覆盖的关系，谁在前面谁work，好比"//abc/*"，那么此时后面的块注释起始符被忽略掉，一样"/*abc//"，后面的单行注释起始符也不起做用，因此二者之间的先后顺序很重要。博主刚开始想的方法是用string的find函数来分别找"//"和"/*"的起始位置，若是不存在就返回-1，可是须要分多种状况来处理，其是否存在，还有两者的先后顺序，处理起来比较麻烦。起始咱们能够直接按字符来一个一个处理，因为块注释是多行注释，因此一旦以前有了块注释的起始符，当前行的处理方式就有所不一样了，因此咱们须要一个变量blocked来记录当前是否为块注释状态，初始化为false。创建空字符out，用来保存去除注释后的字符。而后咱们遍历整个代码的每一行，遍历每一行中的每个字符，若是当前字符是最后一个字符了，说明不会再有注释了，将当前字符加入out中，不然取出当前位置和下一个位置的两个字符，若是其正好是"/*"，说明以后的部分都是块注释了，咱们将blocked赋值为true，而后指针向后移动一个，明明两个字符啊，为啥只移动一个呢，由于另外一个能够在for循环中的++i移动；若是当前两个字符正好是"//"，说明当前行以后都是注释，咱们并不care后面有啥，因此能够直接break掉当前行；若是都不是，说明当前字符是代码，将其加入out中。好，下面来看blocked为true的状况，说明以后的内容都是块注释的内容，咱们惟一关心的是有没有结束符"*/"，因此仍是先作判断，若是当前不是最后一个字符，说明至少还有两个字符，而后取出两个字符，若是正好是块注释结束符，那么咱们将标识重置为false，指针要后移动一个。当前行遍历完后，若是out不为空，且blocked为false，则将out存入结果res中，参见代码以下：

 

class Solution {
public:
    vector<string> removeComments(vector<string>& source) {
        vector<string> res;
        bool blocked = false;
        string out = "";
        for (string line : source) {
            for (int i = 0; i < line.size(); ++i) {
                if (!blocked) {
                    if (i == line.size() - 1) out += line[i];
                    else {
                        string t = line.substr(i, 2);
                        if (t == "/*") blocked = true, ++i;
                        else if (t == "//") break;
                        else out += line[i];
                    }
                } else {
                    if (i < line.size() - 1) {
                        string t = line.substr(i, 2);
                        if (t == "*/") blocked = false, ++i;
                    }
                }
            }
            if (!out.empty() && !blocked) {
                res.push_back(out);
                out = "";
            }
        }
        return res;
    }
};

 

相似题目：

Ternary Expression Parser

Mini Parser

 

参考资料：

https://discuss.leetcode.com/topic/109943/c-easy-solution

https://discuss.leetcode.com/topic/109637/c-o-n-one-pass

 

LeetCode All in One 题目讲解汇总(持续更新中...)