673 - Parentheses Balance

Parentheses Balance

You are given a string consisting of parentheses () and []. A string of this type is said to be correct:

(a)

if it is the empty string

(b)

if A and B are correct, AB is correct,

(c)

if A is correct, (A) and [A] is correct.

Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.

Input 

The file contains a positive integer n and a sequence of n strings of parentheses () and [], one string a line.

Output 

A sequence of Yes or No on the output file.

Sample Input 

3
([])
(([()])))
([()[]()])()

Sample Output 

Yes
No
Yes

————————————————————————————————————————————————————

代码：

#include<iostream>
#include<string.h>
#include<cstdio>
#include<stack>
using namespace std;

int main()
{
char str[128 + 10];
int n;

scanf("%d", &n);
getchar();
while (n--)
{
fgets(str, 128 + 10, stdin); 
int len = strlen(str);
if (str[len - 1] == '\n')
len--;
stack<char> q;
for (int  i = 0; i < len; i++)
{
if (str[i] == '(' || str[i] == '[')
q.push(str[i]);
else
{
if (q.empty())
{
q.push(str[i]);
break;
}
if (q.top() == '(' && str[i] == ')' || 
q.top() == '[' && str[i] == ']')
q.pop();
else
break;
}
}
if (q.empty())
printf("Yes\n");
else
printf("No\n");
}
return 0;
}

分析：

这是一个经典的关于栈的应用的题目——括号配对.第一次提交的时候是 Wrong Answer.再次读题才发现错误所在：

(a)

if it is the empty string

即字符串为空，也是正确的（读题不仔细了），修改后就过了.

这里有个小经验：

注意fgets()和scanf("%s")读入一串字符的区别.fgets()会读入'\n'最为最后一个有效字符,在其后是结束标志'\0'.而scanf,则不会读入'\n',直接是结束标志'\0'。而strlen()是计算的'\0'前的字符数.所以用fgets 和 scanf()读入字符串，用strlen得出的结果往往（注意，我说的是往往，下面会说到例外）是strlen (用fgets读入) = strlen(用scanf("%s")读入) + 1；但有时当不是以'\n'结束时，比如文件末尾没有换行.其结果是一样的.