括号检查器代码的测试用例不正确.对于'(()&

test case incorrect for paranthesis checker code. For #39;(()#39; output sould be #39;not balanced#39; but i#39;m getting #39;balanced#39;(括号检查器代码的测试用例不正确.对于(()输出应该是不平衡但我得到平衡)

问题描述

给定一个表达式字符串 exp.检查 {,"}","(,")","[,"]" 在 exp 中的对和顺序是否正确.例如，程序应该为 exp = [()]{}{[()()]()}" 和 ' 打印 'balanced'不平衡' for exp = [(])"

Given an expression string exp. Examine whether the pairs and the orders of "{","}","(",")","[","]" are correct in exp. For example, the program should print 'balanced' for exp = "[()]{}{[()()]()}" and 'not balanced' for exp = "[(])"

输入

输入的第一行包含一个整数 T，表示测试用例的数量.每个测试用例都由一个表达式字符串组成，单独一行.

The first line of input contains an integer T denoting the number of test cases. Each test case consists of a string of expression, in a separate line.

输出

如果括号是平衡的，则打印 'balanced' 不带引号，否则在单独的行中打印 'notbalanced'.

Print 'balanced' without quotes if the pair of parenthesis is balanced else print 'not balanced' in a separate line.

约束

1 ≤ T ≤ 100
1 ≤ |s| ≤ 105

例子

输入:

3
{([])}
()
([]

输出

balanced
balanced
not balanced

#include <iostream>
#include <cstring>
#include <stack>
using namespace std;

int main() {
    //code
    int t;cin>>t;
    while(t--) {
        string s;
        stack<char> st;
        int i=0,flag =1;
        getline(cin,s);
        int n = s.size();
        while(s[i] != '') {
            if((s[i] == '{') || (s[i] == '[') || (s[i] == '(')) {
                st.push(s[i]);
            }
            else if(!st.empty() && ((st.top() == '{' && s[i] == '}') || 
            (st.top() == '[' && s[i] == ']') || (st.top() == '(' && s[i] == ')'))) st.pop();
            else {
                flag = 0;
                break;
            }
            i++;
        }
        (st.empty() && flag) ? cout<<"balanced
" : cout<<"not balanced
";
    }
    return 0;
}

推荐答案

用调试器单步执行你的代码，第一次循环你的字符串是空的.

Stepping through your code with a debugger, the first time through the loop your string is empty.

这是因为 cin >>t 在第一行的换行符处停止读取，第一行 getline(cin, s) 收到一个空字符串.

This is because cin >> t stops reading at the newline character of the first line, and the first getline(cin, s) line receives an empty string.

在 cin >> 之后直接调用 getline(cin, dummy_string)t 读取该行的其余部分，或切换循环以使用 cin >>s 代替.

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