| 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
Miguel Revilla
2000-08-14
题意： 如果有一个‘（‘那么就要有一个）与它匹配，如果有一个‘[‘那么就要有一个‘]‘与它匹配.。如果都满足，输出”YES“，else 输出”NO“. 思路； 建立一个栈，左括号’[‘,‘(‘都入栈，右括号‘)‘,‘]‘都出栈依次进行下去，如果栈为空，则满足条件，输出“YES”,如果不满足条件，则输出”NO“。 代码：#include<cstdio>
#include<stack>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
getchar();
while(n--)
{
char a[130];
stack<char> s;
gets(a);
int flag=0;
for(int i=0;a[i]!='\0';i++)
{
if(a[i]=='('||a[i]=='[')
s.push(a[i]);
else if(a[i]==')')
{
if(!s.empty()&&s.top()=='(') s.pop();
else
{flag=1;break;}
}
else if(a[i]==']')
{
if(!s.empty()&&s.top()=='[')
s.pop();
else {flag=1;break;}
}
}
if(flag||!s.empty()) printf("No\n");
else printf("Yes\n");
}
return 0;
}
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时间： 2024-12-23 17:13:30