Java递归实现带括号的四则运算表达式求值

思路： 表达式本身其实就是一种递归的表示方式。对于任何一个表达式(expression)，例如(1/2)8-3(1+1)，我们都可以把它先以+和-为边界分割成几块（term：项）来处理，故分成了(1/2)8和3(1+1)两个部分，对于每个部分，如果含有*或/，再分，分成两块（factor：因子），注意，由于表达式是递归地定义的，于是每个因子里面又可能含有表达式（expression），一直处理到表达式末尾就可以了
Code:

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Calculator {
    
    
    private Queue<Character> input = new LinkedList<Character>();
    private String expString = "(1-3*(3.2-3.1)+7)";//default expression

    public Calculator () {
    
    
        init();
    }
    public Calculator (String exp) {
    
    
        expString = exp;
        init();
    }
    public double getResult () {
    
    
        return expression_value();
    }
    private void init () {
    
    
        for (int i=0;i<expString.length();i++) {
    
    
            input.offer(expString.charAt(i));
        }
    }
    private double expression_value () {
    
    //there maybe term in an expression
        double res = term_value();
        while (true && !input.isEmpty()) {
    
    
            char op = input.peek();
            if (op == '+' || op == '-') {
    
    
                input.poll();
                double val = term_value();
                if (op == '+')
                    res += val;
                else
                    res -= val;
            } else {
    
    
                break;
            }
        }
        return res;
    }
    private double term_value () {
    
    //there maybe factor in a term
        double res = factor_value();
        while (true && !input.isEmpty()) {
    
    
            char c = input.peek();
            
            if (c == '*' || c == '/') {
    
    
                input.poll();
                double val = factor_value();
                if (c == '*')
                    res *= val;
                else
                    res /= val;
            } else {
    
    
                break;
            }
        }
        return res;
    }
    private double factor_value () {
    
    //there maybe expression in a factor
        double res = 0;
        char c = input.peek();
        if (c == '(') {
    
    
            input.poll();
            res = expression_value();
            input.poll();
        } else {
    
    
            while (Character.isDigit(c)) {
    
    
                res = res * 10 + c - '0';
                input.poll();
                c = input.peek();
            }
            if (c == '.') {
    
    
                input.poll();
                c = input.peek();
                double t = 0.1;
                while (Character.isDigit(c)) {
    
    
                    res += (c - '0') * t;
                    t *= 0.1;
                    input.poll();
                    c = input.peek();
                }
            }
        }
        return res;
    }
    public static void main(String[] args) {
    
    
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        Calculator cal = new Calculator(s);
        System.out.println(cal.getResult());
    }
}

ps: 参考了博客传送门里面的的C++版代码