Java字符串型方程两位数值问题

Question

我被分配创建一个 Java 程序，它读取一个字符串方程并计算它的值。但是到目前为止，我设法创建的代码只能计算个位数的整数。而且我不知道如何使更长的整数起作用。为了处理两位甚至三位数的整数，我的函数应该是什么样的？

public class Ex8 {
    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here

    String e =  "1 - 3 * 18 / 4 + 2";
    double t = executeStringEquation(e);
    System.out.println(e + " = "+t);
    }



      public static double executeStringEquation(String inputEquation)
      {
        if("".equals(inputEquation))
        {
            return 0;
        }
        else if(inputEquation.length() == 1)
        {
            return Integer.valueOf(inputEquation);
        }

        else
        {
            String _a = String.valueOf(inputEquation.charAt(0));
            String _b = String.valueOf(inputEquation.charAt(1));
            if("+".equals(_b) || "-".equals(_b) )
            {
                if("+".equals(_b))
                {
                    return Integer.valueOf(_a) + executeStringEquation(inputEquation.substring(2));
                }
                else
                {// "-"
                    return Integer.valueOf(_a) - executeStringEquation(inputEquation.substring(2));
                }
            }
            else
            {// "*" or "/"
                boolean isMulti = ("*".equals(_b));
                String  _c = String.valueOf(inputEquation.charAt(2));                
                Integer tmp = 0;
                if(isMulti)
                {
                    tmp = Integer.valueOf(_a) * Integer.valueOf(_c);
                }
                else
                {
                    tmp = Integer.valueOf(_a) / Integer.valueOf(_c);
                }
                String newStr = String.valueOf(tmp) + inputEquation.substring(3);                
                return executeStringEquation(newStr);
            }
        }
    }
}

Answer 1

我正在开发一个函数来使用多位数字的后缀评估来评估您的表达式。这样做的想法是您可以在此处阅读的后缀和中缀表示法：http://interactivepython.org/runestone/static/pythonds/BasicDS/InfixPrefixandPostfixExpressions.html

这是此任务的一段代码。 我更正了表达式评估器的错误，现在评估是正确的，请检查一下：

package com.stackoverflow.infix;

import java.util.Collections;
import java.util.LinkedList;
import java.util.List;
import java.util.Stack;
import java.util.StringTokenizer;

public class InfixPostfixEvaluator {

    private static final String operators = "-+/*";

    private int getPrecedence(String input) {
        char operator = input.charAt(0);
        int ret = 0;
        if (operator == '-' || operator == '+') {
            ret = 1;
        } else if (operator == '*' || operator == '/') {
            ret = 2;
        }
        return ret;
    }

    private boolean operatorGreaterOrEqual(String op1, String op2) {
        return getPrecedence(op1) >= getPrecedence(op2);
    }

    private boolean isOperator(String operator) {
        return operators.indexOf(operator) >= 0;
    }

    private boolean isOperand(String val) {
        try {
            Double.parseDouble(val);
        } catch (NumberFormatException nfe) {
            return false;
        }
        return true;
    }

    public Stack<String> convert2Postfix(List<String> expressionInTokens) {
        Stack<String> stack = new Stack<String>();
        Stack<String> postExpression = new Stack<String>();

        for (String c : expressionInTokens) {
            if (isOperator(c)) {
                while (!stack.isEmpty()) {
                    if (operatorGreaterOrEqual(stack.peek(), c)) {
                        postExpression.push(stack.pop());
                    } else {
                        break;
                    }
                }
                stack.push(c);
            } else if (isOperand(c)) {
                postExpression.push(c);
            }
        }
        while(!stack.isEmpty()){
            postExpression.push(stack.pop());
        }
        Collections.reverse(postExpression);
        return postExpression;
    }

    public int evaluatePostfix(Stack<String> expression) {
        Stack<Integer> stack = new Stack<Integer>();
        while (!expression.isEmpty()) {
            String c = expression.pop();
            if (isOperand(String.valueOf(c))) {
                stack.push(Integer.parseInt(c));
            } else if (isOperator(String.valueOf(c))) {
                int op1 = stack.pop();
                int op2 = stack.pop();
                int result;
                switch (c) {
                case "*":
                    result = op1 * op2;
                    stack.push(result);
                    break;
                case "/":
                    result = op2 / op1;
                    stack.push(result);
                    break;
                case "+":
                    result = op1 + op2;
                    stack.push(result);
                    break;
                case "-":
                    result = op2 - op1;
                    stack.push(result);
                    break;
                }
            }
        }
        return stack.pop();
    }

    /**
     * Takes an expression and parses it, returning a list with every element of
     * the expression.The expression must not have white spaces, or characters
     * different from numbers or the operands + - / *
     * 
     */
    public List<String> split(String expression) {
        List<String> a = new LinkedList<String>();

        StringTokenizer st = new StringTokenizer(expression, "+*/-", true);
        while (st.hasMoreTokens())
            a.add(st.nextToken());
        return a;
    }

}

还有测试

package com.stackoverflow.infix;

import static org.junit.Assert.assertEquals;

import java.util.LinkedList;
import java.util.List;
import java.util.Stack;
import java.util.StringTokenizer;

import org.junit.Test;

public class TestInfixPostfixEvaluator {

    @Test
    public void testWhole() {

        InfixPostfixEvaluator eval = new InfixPostfixEvaluator();

        String infixExpr = "5+50-10";
        List<String> expression = eval.split(infixExpr);
        Stack<String> stack = eval.convert2Postfix(expression);
        assertEquals(45, eval.evaluatePostfix(stack));

        infixExpr = "2-3+4";
        expression = eval.split(infixExpr);
        stack = eval.convert2Postfix(expression);
        assertEquals(3, eval.evaluatePostfix(stack));

        infixExpr = "305-303+4";
        expression = eval.split(infixExpr);
        stack = eval.convert2Postfix(expression);
        assertEquals(6, eval.evaluatePostfix(stack));

        infixExpr = "2+3";
        expression = eval.split(infixExpr);
        stack = eval.convert2Postfix(expression);
        assertEquals(5, eval.evaluatePostfix(stack));

        infixExpr = "2*5+5";
        expression = eval.split(infixExpr);
        stack = eval.convert2Postfix(expression);
        assertEquals(15, eval.evaluatePostfix(stack));

        infixExpr = "2*5/5*5";
        expression = eval.split(infixExpr);
        stack = eval.convert2Postfix(expression);
        assertEquals(10, eval.evaluatePostfix(stack));

        infixExpr = "4+2*16/4";
        expression = eval.split(infixExpr);
        stack = eval.convert2Postfix(expression);
        assertEquals(12, eval.evaluatePostfix(stack));

        infixExpr = "16*3/4+2";
        expression = eval.split(infixExpr);
        stack = eval.convert2Postfix(expression);
        assertEquals(14, eval.evaluatePostfix(stack));

    }

    @Test
    public void testSplit() {
        String exp = "15+20+84/205-1";
        LinkedList<String> a = new LinkedList<String>();

        StringTokenizer st = new StringTokenizer(exp, "+*/-", true);
        while (st.hasMoreTokens())
            a.add(st.nextToken());
        System.out.println(a.toString());
    }

}

Answer 2

考虑读取整个字符串并在空白处拆分。这可能不是一个有效的解析规则，但它可以让您获得多位整数。