递归解析计算器java

Question

我正在尝试在 Java 中创建一个用于加法、乘法和阶乘的递归解析计算器，但我正在努力阅读用户输入以将输入拆分为数字和运算符的第一部分。在调试时，我试图查看哪里出了问题，我发现当“+”通过 if else 语句时，它只是跳过了它。我真的不确定问题是什么，我最初尝试使用标记，然后拆分为子字符串，但当时也不太顺利。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

我注意到您使用的是java.util.Scanner。我编写了一个脚本，应该按照您的所有标准为您完成任务：

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);
    }
}

使用示例：对于二元运算符，输入+21，它会为您添加它们。对于一元，输入 !3，它将产生阶乘。现在，您可以使用一元和二元运算符尝试任何数字组合和排列链，它会递归地为您计算值。

例如，考虑输入 !*3+12：它将添加 1 和 2，然后将其乘以 3 最后，它从整个表达式中计算出阶乘，从而得到预期的362880。

Answer 2

为什么不把输入的计算字符串赋值给一个字符数组，遍历数组，匹配字符'+'、'-'、'*'？