c++ - 为什么我的C++程序崩溃了。我是圆括号的实现问题

Question

我正在对黑客等级实施平衡括号问题。 link
当我在hackerrank ide上运行我的程序时，它显示以下警告。

Solution.cpp: In function ‘std::__cxx11::string isBalanced(std::__cxx11::string)’:
Solution.cpp:51:17: warning: this ‘if’ clause does not guard... [-Wmisleading-indentation]
                 if('{'!=s[i])
                 ^~
Solution.cpp:54:21: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the ‘if’
                     bracket.pop();
                     ^~~~~~~
Solution.cpp:25:16: error: control reaches end of non-void function [-Werror=return-type]
     stack<char>bracket;
                ^~~~~~~
cc1plus: some warnings being treated as errors

当我在本地计算机上运行程序时，读取整数(test case)后，程序崩溃。
为了简化代码，我添加了注释。
问题在isBalanced()函数中显示。
我的代码:

#include <bits/stdc++.h>

using namespace std;

// return true if the passed character belongs to ar[], false otherwise
bool find(char ch,char ar[]){
    for(int i=0;i<3;i++)
        if(ar[i]==ch)
            return true;

    return false;
}


string isBalanced(string s) {
    int size=s.length();
    stack<char>bracket;
    char open[3]={'{','[','('};

    // if size of string is odd then expression is not balanced
    if(size%2!=0)
        return "NO";

    for(int i=0;i<size;i++){
    // if current bracket is opening bracket
    // then push it into bracket stack
    if(find(s[i],open)){
        bracket.push(s[i]);
    }
    // if current bracket is closing bracket
    // then match top element of bracket stack with current bracket
    // if both are matched then pop an element from bracket stack
    // if both are not matched return "NO"
    else{
        // if there is an element to match and current stack is empty then expression is not balanced
        if(bracket.empty())
        return "NO";

        else
        switch (bracket.top()) {
         case '{': 

                if('{'!=s[i])
                 return "NO";

                 bracket.pop();
             break;

        case '(':
        if('('!=s[i])
            return "NO";

        bracket.pop();
        break;

        case '[':
        if('['!=s[i])
            return "NO";

        bracket.pop();
        break;
        }

        bracket.pop();
    }
    return "YES";
}

}

int main()
{
    int t;
    cin >> t;
    for (int t_itr = 0; t_itr < t; t_itr++) {
        string s;
        getline(cin, s);
        string result = isBalanced(s);
        cout << result << "\n";
    }
    return 0;
}

Answer 1

两件事情:

不要混合使用getline和运算符>>。要么这样做:

int main()
{
    int t;
    cin >> t;
    cin.ignore();
    for (int t_itr = 0; t_itr < t; t_itr++) {
        string s;
        getline(cin, s);
        string result = isBalanced(s);
        cout << result << "\n";
    }
    return 0;
}

或这样做:

int main()
{
    int t;
    cin >> t;
    for (int t_itr = 0; t_itr < t; t_itr++) {
        string s;
        cin >> s;
        string result = isBalanced(s);
        cout << result << "\n";
    }
    return 0;
}

视情况而定。

您的逻辑有缺陷。这是isBalanced()的正确实现:

string isBalanced(string s) {
    int size=s.length();
    stack<char>bracket;
    char open[3]={'{','[','('};

    if(size%2!=0)
        return "NO";

    for(int i=0;i<size;i++){
        cout << i << endl;
        if(find(s[i],open)){
            bracket.push(s[i]);
        }
        else{
            if(bracket.empty())
                return "NO";
            else
            {
                switch (bracket.top()) {
                    case '{': 
                        if('}'!=s[i])
                            return "NO";
                        break;

                    case '(':
                        if(')'!=s[i])
                            return "NO";
                        break;

                    case '[':
                        if(']'!=s[i])
                            return "NO";
                    break;
                }
                bracket.pop();
            }
        }
    }
    return bracket.empty() ? "YES" : "NO";
}

建议:请始终使用适当的缩进。