java写一个时间复杂度低的带括号计算器

好的，我可以帮你回答这个问题。下面是一个使用栈来实现带括号的计算器的 Java 代码，时间复杂度为 O(n)。

import java.util.Stack;
import java.util.Scanner;

public class Calculator {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String expression = sc.nextLine();
        System.out.println("结果为：" + evaluate(expression));
    }

    public static int evaluate(String expression) {
        Stack<Integer> numbers = new Stack<>();
        Stack<Character> operators = new Stack<>();

        for (int i = 0; i < expression.length(); i++) {
            char c = expression.charAt(i);
            if (c >= '0' && c <= '9') {
                int number = c - '0';
                while (++i < expression.length() && expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {
                    number = number * 10 + (expression.charAt(i) - '0');
                }
                i--;
                numbers.push(number);
            } else if (c == '(') {
                operators.push(c);
            } else if (c == ')') {
                while (operators.peek() != '(') {
                    numbers.push(apply(operators.pop(), numbers.pop(), numbers.pop()));
                }
                operators.pop();
            } else if (c == '+' || c == '-' || c == '*' || c == '/') {
                while (!operators.empty() && hasPrecedence(c, operators.peek())) {
                    numbers.push(apply(operators.pop(), numbers.pop(), numbers.pop()));
                }
                operators.push(c);
            }
        }

        while (!operators.empty()) {
            numbers.push(apply(operators.pop(), numbers.pop(), numbers.pop()));
        }

        return numbers.pop();
    }

    public static int apply(char operator, int b, int a) {
        switch (operator) {
            case '+':
                return a + b;
            case '-':
                return a - b;
            case '*':
                return a * b;
            case '/':
                if (b == 0) throw new ArithmeticException("除数不能为零");
                return a / b;
            default:
                return 0;
        }
    }

    public static boolean hasPrecedence(char op1, char op2) {
        if (op2 == '(' || op2 == ')') return false;
        if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-')) return false;
        return true;
    }
}

可以使用该计算器进行除法、加法、减法和乘法，同时支持括号运算。你可以输入这样的表达式进行测试：

2 * (3 + 4) - 5 / 2

该表达式的结果应该是 12。