递归解析计算器java

Question

我试图用java为加，乘和阶乘创建一个递归解析计算器，但是我在阅读用户输入以将输入分为数字和运算符的第一部分就很挣扎。 在调试时，我试图查看错误的出处，并且发现当“ +”通过if else语句时，它只是跳过了它。 我真的不确定这个问题是什么，我最初尝试使用令牌，然后拆分为子字符串，但是那时也不顺利。 任何帮助，将不胜感激。 谢谢

package com.company;
import java.util.Scanner;

class Main {

    public static void main(String[] param) {
        String input = input("Please enter an expression");
        int n = input.length()-1;
        String[] splitter = input.split("(?<=\\G.)");
        split(input, n);
        //int result = calculate(input);
        //String[] splitter = input.split("(?<=\\G.)");
    }
    public static String split(String input, int n) {
        String[] splitter = input.split("(?<=\\G.)");
        System.out.println(splitter[n]);
        String symbol = splitter[n];
        if (symbol.equals("+")) {
            evalADD(n, splitter);
        }
        if (symbol.equals("*")) {
            evalMULT(n, splitter);
        }
        if (symbol.equals("!")) {
            evalFACT(n, splitter);
        }
        else if (Integer.parseInt(splitter[n]) >= 0 && Integer.parseInt(splitter[n]) <=9)
        {
            if (n != 0) {
                n = n - 1;
                split(input, n);
            }
        }

        if (n != 0)
            n = n - 1;
        split(input, n);
        return input;
    }
    public static int evalADD(int n, String [] splitter){
        int arg1;
        int arg2;
        int result;
        arg1 = Integer.parseInt(splitter[n+1]);
        arg2 = Integer.parseInt(splitter[n+2]);
        result = arg1 + arg2;
        return result;
    }
    public static int evalMULT(int n, String [] splitter){
        int arg1;
        int arg2;
        int result;
        arg1 = Integer.parseInt(splitter[n+1]);
        arg2 = Integer.parseInt(splitter[n+2]);
        result = arg1 * arg2;
        return result;
    }
    public static int evalFACT(int n, String [] splitter){
        int arg1;
        int arg2;
        int result;
        arg1 = Integer.parseInt(splitter[n+1]);
        arg2 = Integer.parseInt(splitter[n+2]);
        result = arg1 - arg2;
        return result;
    }
    public static String input(String message) {
        Scanner scanner = new Scanner(System.in);
        System.out.println(message);
        return (scanner.nextLine());
    }
}

Answer 1

为什么不将输入的计算字符串分配给字符数组并遍历该数组并匹配字符'+'，'-'，'*'？

Answer 2

我注意到您正在使用java.util.Scanner 。 我编写了一个脚本，该脚本应按照您的所有条件为您完成任务：

import java.util.Scanner;

class recursiveParsingCalculator {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        // Ask user to input the expression
        System.out.println("Please input the expression");
        String userInput = scanner.nextLine();
        System.out.println(
                "And the final result is: " + recursiveCalculation(userInput, userInput.length() - 1, 0, 0, 0));
        scanner.close();
        System.exit(0);
    }

    // Identify the type of character at a specific position
    public static char charOfString(String userInput, int i) {
        return userInput.charAt(i);
    }

    /*
     * Position must be userInput.length() - 1 initially. currentResults, operand1
     * and operand2 are also meant to be initilized with 0.
     */
    public static int recursiveCalculation(String userInput, int position, int operand1, int operand2,
            int currentResults) {
        // If position is zero, just output the operand.
        if (position == 0) {
            if (Character.isDigit(charOfString(userInput, position))) {
                return charOfString(userInput, position) - '0';
            } else {
                System.out.println("Invalid input.");
            }
        }
        if (position > -1) {
            // Check if it is a number or an operator
            if (Character.isDigit(charOfString(userInput, position))) {
                operand1 = charOfString(userInput, position) - '0'; // First operand

                // Check if 2nd char is a number or an operator.
                if (Character.isDigit(charOfString(userInput, position - 1))) {
                    operand2 = charOfString(userInput, position - 1) - '0';
                    position = position - 1;
                }
            } else {
                // If it is an operator, then proceed to compute the results so far
                char operator = charOfString(userInput, position);

                // If it is a binary situation
                if (operator == '+' || operator == '*') {
                    currentResults = binaryOperator(operator, operand1, operand2);
                    operand2 = currentResults;
                }
                // If it is an unary situation
                else if (operator == '!') {
                    if (currentResults == 0) {
                        currentResults = operand1;
                    }
                    currentResults = unaryOperator(currentResults);
                    operand2 = currentResults;
                } else {
                    System.out.println("Invalid operator");
                    return 0; // Return zero by default
                }
            }
            position = position - 1;
        }
        if (position > -1) {
            return recursiveCalculation(userInput, position, operand1, operand2, currentResults);
        } else {
            return currentResults;
        }
    }

    public static int binaryOperator(char operator, int operand1, int operand2) {
        switch (operator) {
        case '+':
            return operand1 + operand2;
        case '*':
            return operand1 * operand2;
        default:
            System.out.println("Invalid binary Operator");
            return 0; // Return zero by default
        }
    }

    // Calculate the factorial
    public static int unaryOperator(int operand) {
        if (operand <= 1)
            return 1;
        else
            return operand * unaryOperator(operand - 1);
    }
}

用法示例：对于二进制运算符，输入+21 ，它将为您添加它们。 对于一元，输入！3 ，它将产生阶乘。 现在，您可以尝试使用一元和二进制运算符对数字进行组合和排列的任何链，它将为您递归计算值。

例如，考虑输入！* 3 + 12 ：它将输入1和2 ，然后将其乘以3 ，最后，它将计算整个表达式中的阶乘，从而得到362880 。