求解表达式的简单 Java 计算器

Question

我正在研究求解表达式的计算器。我想弄清楚如何让它按 PEMDAS 顺序计算。我有一个 for 循环来遍历数组列表和一个调用 class 进行数学运算的开关。我已经尝试过 if 语句，但一直无法弄清楚。

我如何更改它以确保表达式以正确的顺序求解？这是我目前所拥有的：

/*
Peter Harmazinski
Simple Calculator

This program solves expressions
*/

import java.util.*;

public class SimpleCalculator2 {
    static SimpleMath math = new SimpleMath();

    public static void main(String[] args) {
        Scanner console = new Scanner(System.in);
        boolean again = true;
        double number1 = 0.0;
        double number2 = 0.0;
        double answer = 0.0;
        double results = 0.0;
        String delims = "[ ]+";

        getIntroduction();

        while (again) {
            System.out.println("Please enter your expression.");
            String input = console.nextLine();
            System.out.println("This is the user's input: " + input);

            //Parses string into array list
            List<String> list = new ArrayList<String>(Arrays.asList(input.split(delims)));
            System.out.println("list: " + list);

            results = doMath(list, number1, number2);
            getResults(results);
        }
        console.close();
    }

    public static void getIntroduction() {
        System.out.println("This is a simple calculator that solves expressions.");
    }

    //Traverses array list to identify operators and does math for surrounding numbers
    //then answer is inserted in i-1 element and the elements i and i+1 are deleted

    public static double doMath(List<String> list, double number1, double number2) {
        double answer = 0.0;
        double results = 0.0;
        while (list.size() > 1) {
            for (int i = 0; i < list.size(); i++) {
                    switch (list.get(i)) {
                        case "*" :
                            number1 = Double.parseDouble(list.get(i - 1));
                            number2 = Double.parseDouble(list.get(i + 1));
                            answer = math.multiply(number1, number2);
                            System.out.println(answer);
                            list.add(i - 1, Double.toString(answer));
                            list.subList(i, i + 3).clear();
                            System.out.println(list);
                            break;
                        case "/" :
                            number1 = Double.parseDouble(list.get(i - 1));
                            number2 = Double.parseDouble(list.get(i + 1));
                            answer = math.divide(number1, number2);
                            System.out.println(answer);
                            list.add(i - 1, Double.toString(answer));
                            list.subList(i, i + 3).clear();
                            System.out.println(list);
                            break;
                        case "+" :
                            number1 = Double.parseDouble(list.get(i - 1));
                            number2 = Double.parseDouble(list.get(i + 1));
                            answer = math.add(number1, number2);
                            System.out.println(answer);
                            list.add(i - 1, Double.toString(answer));
                            list.subList(i, i + 3).clear();
                            System.out.println(list);
                            break;
                        case "-" :
                            number1 = Double.parseDouble(list.get(i - 1));
                            number2 = Double.parseDouble(list.get(i + 1));
                            answer = math.subtract(number1, number2);
                            System.out.println(answer);
                            list.add(i - 1, Double.toString(answer));
                            list.subList(i, i + 3).clear();
                            System.out.println(list);
                            break;
                        }
                }   
            }   
        return answer;
    }

    public static void getResults(double results) {
        System.out.println("Results are: " + results);
    }
}

Answer 1

我认为实现此目的的一个非常标准的算法是 Dijkstra 的调车场算法，然后是后缀评估。你可以在这里阅读它，伪代码在这里，但你可能需要对基本数据结构有一些了解： http://en.wikipedia.org/wiki/Shunting-yard_algorithm 。

如果您不了解堆栈、队列和后缀表示法，您可以编写一个更慢、更直接但更混乱的实现。我也用 sin、cos、trig、log 尝试过一次，结果代码运行正常，但我不会再试一次。

基本上，我们的想法是只用 1 个运算符找到最高优先级的表达式，对其求值，然后用它替换。这是一些伪代码：

input = [user input]
while expression still contains (, ), +, -, *, or /:
    toEvaluate = highest priority expression with 1 operator (e.g. 1+2, or 2*3)
    calculate the decimal value of toEvaluate
    modify input so that you replace toEvaluate with its decimal value

请注意，在您的 doMath() 实现中，在您的 for 循环中，您只需在看到所有运算符时立即对其进行评估。例如，考虑

1+2*3

您将首先看到 +，计算 1+2，然后将结果乘以 3。相反，您需要先遍历整个列表，找到优先级最高的运算符，对其求值，然后从头开始再次.

Answer 2

既然你要将它输入到一个字符串中，为什么不制作一个遍历每个字符的 for 循环？

// This for loop will get you the string with only one operator in it.
// EX: 30+32
// You will have to find some way to compute this string.
for(int i = 0; i < userInput.length(); i++)
{
   //int a will be the INDEX of the first operator
   //int b will be the INDEX of the place where a new operator comes up 
   char c = userInput.getCharAt(i);
   a = 0;
   //I'm just gonna do it with + and - for now. Add any other operators.
   if(c == '+' || c == '-')
   {
       b = i;
       String stringToCompute = userInput.substring(a,b);
       //Find some way to take the string there then compute it. 
       //Maybe another for loop with stringToCompute to find where the
       // operator is and then add/subtract the two doubles.  

       // Now reset it 
       a = i;
       b = null;
   }
}

Answer 3

这里有几个可能的设计来解决这个问题：

您可以将处理分为两部分：解析和求值。在解析阶段，您将字符串转换为表示将如何对其进行评估的数据结构。在评估阶段，您遍历树并对表达式进行评估。
您可以按运算符拆分列表以评估最后（加号和减号），评估每个段（乘法和除法）。如果设计得当，这可能是递归的。

第一个选项更好，因为它更可扩展，但实施起来也更费力。这是通过解析和评估生成的示例数据结构：

interface Term {
    double getValue();
}

enum Operator {
    MULTIPLY, DIVIDE, ADD, SUBTRACT;
    double getValue(List<Double> operands) {
        ...
    }
}

class Operation implements Term {
    List<Term> operands;
    Operator operator;
    double getValue() {
        return operator.getValue(operands);
    }
}

class Constant implements Term {
    private final double value;
    double getValue() {
        return value;
    }
}

求解表达式的简单 Java 计算器

Simple Java Calculator to Solve Expressions

java

loops

arraylist

switch-statement

calculator