1978:扩号匹配问题(做法1：闭门造车，做法2：作者代码)

做法1：闭门造车

#include <iostream>
#include <cstdio>
#include <string>
#include <stack>

using namespace std;

const int MAXSIZE=100;

stack<int> Left;

int main()
{
    string str;

    while(cin>>str)
    {
        int arr[MAXSIZE]= {0};

        for(int i=0; i<str.size(); ++i)
        {
            if(str[i]=='(')
            {
                Left.push(i);
            }
            if(str[i]==')')
            {
                if(!Left.empty())
                {
                    Left.pop();
                }
                else
                {
                    arr[i]=1;
                }
            }
        }
        while(!Left.empty())
        {
            arr[Left.top()]=-1;
            Left.pop();
        }
        cout<<str<<endl;
        for(int i=0; i<str.size(); ++i)
        {
            if(arr[i]==1)
            {
                cout<<'?';
            }
            else if(arr[i]==-1)
            {
                cout<<'$';
            }
            else
            {
                cout<<' ';
            }
        }
        cout<<endl;
    }
}

做法2：作者代码
作者链接
https://github.com/BenedictYoung/Lecture/blob/master/Chapter5/5.5.cpp

/*
* 题目名称：扩号匹配问题
* 题目来源：OpenJudge 1978
* 题目链接：http://ccnu.openjudge.cn/practice/1978/
* 代码作者：杨泽邦(炉灰)
*/

#include <iostream>
#include <cstdio>
#include <string>
#include <stack>

using namespace std;

int main() {
    string str;
    while (cin >> str) {
        stack<int> brackets;
        string answer(str.size(), ' ');         //初始化为输入长度个空格
        for (int i = 0; i < str.size(); ++i) {
            if (str[i] == '(') {
                brackets.push(i);               //压入左括号下标
            } else if (str[i] == ')') {
                if (!brackets.empty()) {
                    brackets.pop();
                } else {
                    answer[i] = '?';            //右括号不匹配
                }
            }
        }
        while (!brackets.empty()) {
            answer[brackets.top()] = '$';       //左括号不匹配
            brackets.pop();
        }
        cout << str << endl;
        cout << answer << endl;
    }
    return 0;
}