All in one 1353 -- expression bracket matching (stack)

Topic website http://ybt.ssoier.cn:8088/problem_show.php?pid=1353

【 Title Description 】

Let's say an expression has letters （ A lowercase letter ）、 Operator （+,-,∗,/） And left and right little （ round ） Bracket construction , With “@” As the terminator of an expression . Please write a program to check whether the left and right parentheses in the expression match , If the match , Then return to “YES”; Otherwise return to “NO”. The length of the expression is less than 255, The left parenthesis is less than 20 individual .

【 Input 】

A line of data , That's the expression .

【 Output 】

a line , namely “YES” or “NO”.

【 sample input 】

2*(x+y)/(1-x)@

【 sample output 】

YES

【 Tips 】

【 The sample input 2】

(25+x)*(a*(a+b+b)@

【 Sample output 2】

NO

Examine the subject

Expression has no spaces , direct cin that will do . Numbers , Symbol , Letters have no effect on the title

Ideas

1. Judge each character in turn

2. If the current character is @, Exit loop

3. If the current character is an open parenthesis , Push

4. If it's right bracket , Determine whether there is an open bracket in the stack . without , Note that there is no left parenthesis to match it , Output NO; Otherwise, let an open bracket come out of the stack , Indicates that it has been paired

for(int i=0;i<str.size();i++)
{
	if(str[i]=='@') break;
	if(str[i]=='(')stk.push(str[i]); 
	if(str[i]==')'){
		if(stk.empty())
		{
			cout<<"NO";
			return 0;
		}
		stk.pop();
	}
}

5. After the loop ends, judge whether the stack is empty . If it's not empty , Note that there are still left parentheses that have not been matched , Output NO, Otherwise output YES

if(stk.empty()) {
	cout<<"YES";
}
else{
	cout<<"NO";
	return 0;
}

Complete code

#include<bits/stdc++.h>
using namespace std;
string str;
stack<char>stk;
int main()
{
	cin>>str;
	for(int i=0;i<str.size();i++)
	{
		if(str[i]=='@') break;
		if(str[i]=='(')stk.push(str[i]); 
		if(str[i]==')'){
			if(stk.empty())
			{
				cout<<"NO";
				return 0;
			}
			stk.pop();
		}
	}
	if(stk.empty()) {
		cout<<"YES";
	}
	else
	cout<<"NO";
	return 0;
}