[LeetCode] Remove Comments 移除註釋

 

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.

 

這道題讓咱們移除代碼中的註釋部分，就是寫代碼中常常遇到的兩種註釋，單行註釋和多行註釋，也能夠叫塊註釋，固然最最重要的就是要找到這兩種註釋的起始標識符"//"和"/*"，注意它們二者之間存在覆蓋的關係，誰在前面誰work，好比"//abc/*"，那麼此時後面的塊註釋起始符被忽略掉，一樣"/*abc//"，後面的單行註釋起始符也不起做用，因此二者之間的先後順序很重要。博主剛開始想的方法是用string的find函數來分別找"//"和"/*"的起始位置，若是不存在就返回-1，可是須要分多種狀況來處理，其是否存在，還有兩者的先後順序，處理起來比較麻煩。起始咱們能夠直接按字符來一個一個處理，因爲塊註釋是多行註釋，因此一旦以前有了塊註釋的起始符，當前行的處理方式就有所不一樣了，因此咱們須要一個變量blocked來記錄當前是否爲塊註釋狀態，初始化爲false。創建空字符out，用來保存去除註釋後的字符。而後咱們遍歷整個代碼的每一行，遍歷每一行中的每個字符，若是當前字符是最後一個字符了，說明不會再有註釋了，將當前字符加入out中，不然取出當前位置和下一個位置的兩個字符，若是其正好是"/*"，說明以後的部分都是塊註釋了，咱們將blocked賦值爲true，而後指針向後移動一個，明明兩個字符啊，爲啥只移動一個呢，由於另外一個能夠在for循環中的++i移動；若是當前兩個字符正好是"//"，說明當前行以後都是註釋，咱們並不care後面有啥，因此能夠直接break掉當前行；若是都不是，說明當前字符是代碼，將其加入out中。好，下面來看blocked爲true的狀況，說明以後的內容都是塊註釋的內容，咱們惟一關心的是有沒有結束符"*/"，因此仍是先作判斷，若是當前不是最後一個字符，說明至少還有兩個字符，而後取出兩個字符，若是正好是塊註釋結束符，那麼咱們將標識重置爲false，指針要後移動一個。當前行遍歷完後，若是out不爲空，且blocked爲false，則將out存入結果res中，參見代碼以下：

 

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;
    }
};

 

相似題目：

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參考資料：

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

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

 

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