UVa-673-Parentheses Balance

AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 2. Data Structures :: Lists


// 673 - Parentheses Balance
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
using namespace std;
int main(void)
{
	int i, j, n, len;
	char ch[150];
	while(cin >> n)
	{
		getchar();
		for(i=0; i<n; i++)
		{
			stack<char> s;
			cin.getline(ch, 150);
			len = strlen(ch);
			for(j=0; j<len; j++)
			{
				if(ch[j]=='(' || ch[j]=='[')
					s.push(ch[j]);
				else if( ( s.empty() && (ch[j]==')'||ch[j]==']') ) ||
						 ( s.top()=='(' && ch[j]==']' ) ||
						 ( s.top()=='[' && ch[j]==')' ) )
					break;
				else if( ( s.top()=='(' && ch[j]==')' ) ||
						 ( s.top()=='[' && ch[j]==']' ) )
					s.pop();
			}
			if(s.empty() && j==len)
				cout << "Yes" << endl;
			else
				cout << "No" << endl;
		}
	}
	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.
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值