E - 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
错误代码1！
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
int t,sum1,sum2;
cin>>t; //没有考虑不输入任何字符的情况 。当不输入任何字符时，系统要求通过，但此程序却无法通过，因为不输入并没有被考虑，cin也无法识别
for(int i=1;i<=t;i++)
{
char ch[128];
cin>>ch;
int leng=strlen(ch);
int sum1=100,sum2=100;
if((leng+2)%2==0)
{
for(int j=0;j<leng;j++)
{
if(ch[j]=='
') break;
if(ch[j]=='(')sum1++;
if(ch[j]==')')sum1--;
if(ch[j]=='[')sum2++;
if(ch[j]==']')sum2--;
}
if(sum1==100&&sum2==100)
cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
else cout<<"No"<<endl;
}
}
错误代码2！
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
int t,sum1,sum2;
cin>>t;
getchar();
for(int i=1;i<=t;i++)
{
char ch[200];
gets(ch);
int leng=strlen(ch);
int sum1=100,sum2=100;
if(leng==0)cout<<"Yes"<<endl; / /更正了不输入任何字符时输出Yes的情况考虑
else{
if((leng+2)%2==0)
{
for(int j=0;j<leng;j++)
{
if(ch[j]=='(')sum1++;
if(ch[j]==')')sum1--;
if(ch[j]=='[')sum2++;
if(ch[j]==']')sum2--;
}
if(sum1==100&&sum2==100) // 仅仅是数量上的考虑，忽略了题目对“格式”的要求
cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
else cout<<"No"<<endl;
}
}
}
//许多人做了半天题，却发现题目意思完全没有理解
下面解释一下题目要求:
1. ( ) [ ] 的分别个数要相同
2. ( ) [ ]要匹配
例如：（【）】 ：不通过。将（）记做A，[ ] 记做B，原表达式的（)中既不为A，也不为B，不符合题目要求
3.当不输入任何括号时，将输出Yes。
所以从第二点要求来看此题，我们难以再用数组的方法来做此题目
我们选用一种更为简单的方法 : 栈
一个超时的程序！
#include<iostream>
#include<stack>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
int t,i;
char ch[130];
cin>>t;
getchar();
while(t--)
{
stack<int>st;
gets(ch);
int leng=strlen(ch);
for(i=0;i<leng;i++)
{
if(ch[i]=='('){st.push(1);continue;}
if(ch[i]=='['){st.push(2);continue;}
if((ch[i]==')'&&st.top()==1)||(ch[i]==']'&&st.top()==2))st.pop(); else break;
}
if(st.empty()) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
}