递归解析计算器 java

Question

我试图在 java 中创建一个递归解析计算器，用于加法、乘法和阶乘，但我在第一部分苦苦挣扎，只是阅读用户输入以将输入拆分为数字和运营商。在调试时，我试图查看哪里出错了，我发现当“+”通过 if else 语句时，它只是跳过了它。我真的不确定问题出在哪里，我最初尝试使用令牌，然后拆分成子字符串，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

为什么不将输入的计算字符串分配给一个字符数组，然后遍历数组并匹配字符'+'、'-'、'*'？

Answer 2

我注意到您正在使用 java.util.Scanner。我写了一个脚本，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，然后乘以 3，最后计算整个表达式的阶乘，从而得到预期的 362880。

递归解析计算器 java

recursive parsing calculator java

java

recursion

parsing

procedural-programming

calculator