c++ - 我应该使用 map 、3D 数组还是坚持使用堆栈？

Question

到目前为止，除了开大括号多于闭大括号的情况外，我可以使它在每种情况下都能完美地工作。每当堆栈仍然包含任何左括号时，我都不知道如何找到那个特定的左括号。我想知道是否包括 map 或更改我的堆栈以接收 3D(？)数组值以便它可以包含 char 的坐标？如果有更简单的方法，如果你们能帮助指导我实现它，那就太好了，请不要只告诉我答案，因为我不知道我应该如何让它工作。 ;-;

非常感谢!

    void Brackets::check_bracket()
        {
            for (int a = 0; a < line_num; a++)//loops through every line
            {
                for (int c = 0; c < lines[a].length(); c++)//loops through every char in string
                {
                    while (lines[a].find("//") != -1 || lines[a].find("cout") != -1) a++;//checks if there is a comment
                    if (lines[a][c] == '(' || lines[a][c] == '[' || lines[a][c] == '{')//checks if the brace is an open brace
                    {
                        //DBG::cout << "yes" << lines[a][c] << endl;
                        bracket.push(lines[a][c]);//if yes then it pushs the brace in
                    }
                    if (lines[a][c] == ')' || lines[a][c] == ']' || lines[a][c] == '}')//checks if the brace is a close brace
                    {
                        //DBG::cout << bracket.empty() << lines[a][c] << endl;
                        if (bracket.empty())
                        {//if empty then tells user the bracket does not match any open parenthesis
                            //DBG::cout << 1;
                            cout << "closed parenthesis " << lines[a][c] << " does not match any open parenthesis.";
                            return;
                        }
                        //else if the brackets do match an open brace then it pops it out
                        else if ((lines[a][c] == ')' && bracket.top() == '(') || (lines[a][c] == ']' && bracket.top() == '[') || (lines[a][c] == '}' && bracket.top() == '{'))
                        {
                            //DBG::cout << 2;
                            bracket.pop();
                        }//checks if bracket matches any open parenthesis
                        else
                        {
                            //DBG::cout << 3 << a << endl;
                            cout << "Line " << a+1 << ":" << " error at column " << c+1 << ":" << endl;
                            cout << lines[a] << endl;
                            for (int o = 0; o < c; o++)
                                cout << " ";
                            cout << "^" << endl;
                            cout << "closed parenthesis " << lines[a][c] << " does not match open parenthesis " << bracket.top() << " from row " << find_last(a, bracket.top());
                            return;
                        }
                    }
                }
            }
            if (!bracket.empty())//if bracket still contains something in the stack then it finds where the leftover open brace is
            {
                cout << "Unmatched open parenthesis " << bracket.top() << " at row "; //im stuck
            }
            else
            cout << "No parenthesis errors."<<endl;//otherwise no parenthesis errors
        }
        //finds the last problem brace
        string Brackets::find_last(int line, char chk)
        {
            string row_col = "";
            for (int a = line; a >= 0; a--)//loops through lines backwards
            {
                for (int b = lines[a].length() - 1; b >= 0; b--)//loops through length backwards
                {
                    if (lines[a][b] == chk)//checks if the char equals the problem brace
                    {
                        row_col = to_string(a+1) + " and column " + to_string(b+1)+"\n";//returns string of location
                        return row_col;
                    }
                }
            }
            return "";
        }

private:
    string lines[1000];//contains code lines
    int line_num = 0;//number of lines
    stack<char> bracket;//brackets

Answer 1

我只是在堆栈上存储一个简单的结构

struct Bracket {
  char type;
  int line;
  int col;
}

顺便说一句:存储预期的右括号(而不是左括号)可能是有意义的:计算你推的位置很简单，并且会简化当前由 3x && 和 2x 组成的匹配||