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
sourceis 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.
這道題讓我們移除代碼中的注釋部分，就是寫代碼中經常遇到的兩種注釋，單行注釋和多行注釋，也可以叫塊注釋，當然最最重要的就是要找到這兩種注釋的起始標識符"//"和"/*"，注意它們兩者之間存在覆蓋的關系，誰在前面誰work，比如"//abc/*"，那么此時后面的塊注釋起始符被忽略掉，同樣"/*abc//"，后面的單行注釋起始符也不起作用，所以兩者之間的前后順序很重要。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，如果不存在就返回-1，但是需要分多種情況來處理，其是否存在，還有二者的前后順序，處理起來比較麻煩。起始我們可以直接按字符來一個一個處理，由於塊注釋是多行注釋，所以一旦之前有了塊注釋的起始符，當前行的處理方式就有所不同了，所以我們需要一個變量blocked來記錄當前是否為塊注釋狀態，初始化為false。建立空字符out，用來保存去除注釋后的字符。然后我們遍歷整個代碼的每一行，遍歷每一行中的每一個字符，如果當前字符是最后一個字符了，說明不會再有注釋了，將當前字符加入out中，否則取出當前位置和下一個位置的兩個字符，如果其正好是"/*"，說明之后的部分都是塊注釋了，我們將blocked賦值為true，然后指針向后移動一個，明明兩個字符啊，為啥只移動一個呢，因為另一個可以在for循環中的++i移動；如果當前兩個字符正好是"//"，說明當前行之后都是注釋，我們並不care后面有啥，所以可以直接break掉當前行；如果都不是，說明當前字符是代碼，將其加入out中。好，下面來看blocked為true的情況，說明之后的內容都是塊注釋的內容，我們唯一關心的是有沒有結束符"*/"，所以還是先做判斷，如果當前不是最后一個字符，說明至少還有兩個字符，然后取出兩個字符，如果正好是塊注釋結束符，那么我們將標識重置為false，指針要后移動一個。當前行遍歷完后，如果out不為空，且blocked為false，則將out存入結果res中，參見代碼如下：
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; } };
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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