Problem Solving►UVa

UVa 673 - Parentheses Balance

Contents

  1. 1. Problem
  2. 2. Solution
  3. 3. Code

Problem

題目網址
中文網址

Solution

找對應的括號，一遇到 ‘)’ 、 ‘]’ 卻無法有對應的括號，即可停止。

Code

UVa 673
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#include<cstdio>
#include<stack>
#define N 130

int main()
{
    std::stack<char> parentheses;
    int n,i;
    char s[N] = { NULL };
    scanf("%d", &n);

    getchar();
    while (n--)
    {
        fgets(s, N, stdin);

        for (i = 0; s[i] != '\n'; i++)
        {
            if (s[i] == '(' || s[i] == '[')
                parentheses.push(s[i]);
            else if (s[i] == ')')
            {
                if (!parentheses.empty() && parentheses.top() == '(')
                    parentheses.pop();
                else
                    break;
            }
            else if (s[i] == ']')
            {
                if (!parentheses.empty() && parentheses.top() == '[')
                    parentheses.pop();
                else
                    break;
            }
        }

        printf("%s\n", parentheses.empty() && s[i] == '\n' ? "Yes" : "No");

        while (!parentheses.empty())
            parentheses.pop();
    }


    return 0;
}