Parentheses Balance UVA - 673

本文深入探讨了括号匹配问题的解决方法,通过使用栈数据结构实现括号的正确匹配检查,对比两种不同实现方式,指出需注意栈为空时的处理细节,确保算法的完整性和准确性。

//刚开始觉得这道题挺简单的....结果一直总是处理不完结果是No 的数据..(没输出) 。。。。后来发现stack的empty也得加上,不能直接判断栈顶元素与此时数组元素相等,将错误的附在第二个代码上

//AC

#include<iostream>
#include<stack>
#include<string>

using namespace std;

int main()
{
    int n;
    cin>>n;
    string str1;
    getline(cin,str1);
    while(n--)
    {
        string str;
        stack<char> s;
        int i,flag=1;
        getline(cin,str);
        for(i=0;i<str.size();i++)
        {
            if(str[i]=='(')
            {
                s.push(str[i]);
            }
            else if(str[i]=='[')
            {
                s.push(str[i]);
            }
            else if(str[i]==')')
            {
                if(!s.empty() && s.top()=='(')
                {
                    s.pop();
                }

                else
                {

                    flag=0;
                    break;
                }

            }
            else if(str[i]==']')
            {
                if(s.empty() || s.top()!='[')
                {
                    flag=0;
                    break;
                }

                else
                {
                    s.pop();
                }
            }
        }
        if(!s.empty()) flag=0;
        if(flag) cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
    }
    return 0;
}

//RE

#include <iostream>
#include <stack>

using namespace std;

stack<char> a;
stack<char> b;

int main()
{
    int n;

    cin >> n;


    for(int i=0; i<n; i++){
        char t;
        int j=0, flag = 0;

        //while(cin >> t && t != '\n')
        string s;
        cin >> s;

        while(s[j])
        {
            t = s[j];

            if(t == '(' || t == '['){
                a.push(t);
            }
            else if(t == ')') {
                if (a.top() == '(') {
                    a.pop();
                }
                else{
                    flag = 1;
                    break;
                         }
            }

            else if(t == ']'){
                if (a.top() == '['){
                    a.pop();
                }
                else{
                    flag = 1;
                    break;
                }
            }
            j++;

//            cout << j << endl;
        }

//        cout << a.empty() << " " << b.empty() << endl;

        if(flag == 0 && a.empty()){
            cout << "Yes" << endl;
        }
        else
            cout << "No" << endl;
        //清空stack
        while(!a.empty())
            a.pop();
  //      while(!b.empty())
  //          b.pop();
    }
    return 0;
}

 

 

To evaluate a mathematical expression containing integers, '+', '-', and parentheses, with parentheses having precedence over left - to - right evaluation and no multiplication or division, a stack - based approach can be used. The following is a Python implementation: ```python def evaluate_expression(expression): stack = [] num = 0 sign = 1 result = 0 for i in range(len(expression)): char = expression[i] if char.isdigit(): num = num * 10 + int(char) elif char == '+': result += sign * num num = 0 sign = 1 elif char == '-': result += sign * num num = 0 sign = -1 elif char == '(': stack.append(result) stack.append(sign) result = 0 sign = 1 elif char == ')': result += sign * num num = 0 prev_sign = stack.pop() prev_result = stack.pop() result = prev_result + prev_sign * result result += sign * num return result expression = "3 + (4 - 2)" print(evaluate_expression(expression)) ``` In this code: 1. A stack is used to keep track of the previous results and signs when encountering parentheses. 2. When a digit is encountered, it is accumulated to form a multi - digit number. 3. When a '+' or '-' is encountered, the current number is added or subtracted from the result based on the current sign. 4. When a '(' is encountered, the current result and sign are pushed onto the stack, and the result and sign are reset. 5. When a ')' is encountered, the current result is combined with the previous result and sign stored in the stack. Finally, the remaining number and sign are added to the result to get the final evaluation.
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