Question

所以任务是为表达式计算器创建自己的解析器。例如：

输入：3 + 2 * 1-6 / 3 输出：3

输入：3 ++ 2 输出：表达式无效

输入：-5 + 2 输出：-3

输入：5--2 输出：7

这里的代码解决了问题的一部分，除了它有一个固定的输入而负值无法解决，而且我还不确定它是否确实解决了运算符优先级的表达式。但我已经修改它以获取用户的输入表达式。而且我一直想知道如何实现负值的求解。帮助任何人？

没有JAVASCRIPT引擎请。

这是当前的代码

    import java.util.*; 

public class ExpressionParser   
{  
    // Associativity constants for operators  
    private static final int LEFT_ASSOC  = 0;  
    private static final int RIGHT_ASSOC = 1;  

    // Operators  
    private static final Map<String, int[]> OPERATORS = new HashMap<String, int[]>();  
    static   
    {  
        // Map<"token", []{precendence, associativity}>  
        OPERATORS.put("+", new int[] { 0, LEFT_ASSOC });  
        OPERATORS.put("-", new int[] { 0, LEFT_ASSOC });  
        OPERATORS.put("*", new int[] { 5, LEFT_ASSOC });  
        OPERATORS.put("/", new int[] { 5, LEFT_ASSOC });          
    }  

    // Test if token is an operator  
    private static boolean isOperator(String token)   
    {  
        return OPERATORS.containsKey(token);  
    }  

    // Test associativity of operator token  
    private static boolean isAssociative(String token, int type)   
    {  
        if (!isOperator(token))   
        {  
            throw new IllegalArgumentException("Invalid token: " + token);  
        }  

        if (OPERATORS.get(token)[1] == type) {  
            return true;  
        }  
        return false;  
    }  

    // Compare precedence of operators.      
    private static final int cmpPrecedence(String token1, String token2)   
    {  
        if (!isOperator(token1) || !isOperator(token2))   
        {  
            throw new IllegalArgumentException("Invalid tokens: " + token1  
                    + " " + token2);  
        }  
        return OPERATORS.get(token1)[0] - OPERATORS.get(token2)[0];  
    }  

    // Convert infix expression format into reverse Polish notation  
    public static String[] expToRPN(String[] inputTokens)   
    {  
        ArrayList<String> out = new ArrayList<String>();  
        Stack<String> stack = new Stack<String>();  

        // For each token  
        for (String token : inputTokens)   
        {  
            // If token is an operator  
            if (isOperator(token))   
            {    
                // While stack not empty AND stack top element   
                // is an operator  
                while (!stack.empty() && isOperator(stack.peek()))   
                {                      
                    if ((isAssociative(token, LEFT_ASSOC)         && 
                         cmpPrecedence(token, stack.peek()) <= 0) ||   
                        (isAssociative(token, RIGHT_ASSOC)        &&   
                         cmpPrecedence(token, stack.peek()) < 0))   
                    {  
                        out.add(stack.pop());     
                        continue;  
                    }  
                    break;  
                }
                // Push the new operator on the stack  
                stack.push(token);  
            }   
            // If token is a left bracket '('  
            else if (token.equals("("))   
            {  
                stack.push(token);  //   
            }   
            // If token is a right bracket ')'  
            else if (token.equals(")"))   
            {                  
                while (!stack.empty() && !stack.peek().equals("("))   
                {  
                    out.add(stack.pop());   
                }  
                stack.pop();   
            }   
            // If token is a number  
            else   
            {  
            //  if(!isOperator(stack.peek())){
            //      out.add(String.valueOf(token*10));
            //      }
                out.add(token);   
            }  
        }  
        while (!stack.empty())  
        {  
            out.add(stack.pop());   
        }  
        String[] output = new String[out.size()];  
        return out.toArray(output);  
    }  

    public static double RPNtoDouble(String[] tokens)  
    {          
        Stack<String> stack = new Stack<String>();  

        // For each token   
        for (String token : tokens) //for each   
        {  
            // If the token is a value push it onto the stack  
            if (!isOperator(token))   
            {  
                stack.push(token);                  
            }  
            else  
            {          
                // Token is an operator: pop top two entries  
                Double d2 = Double.valueOf( stack.pop() );  
                Double d1 = Double.valueOf( stack.pop() );  

                //Get the result  
                Double result = token.compareTo("*") == 0 ? d1 * d2 :   
                                token.compareTo("/") == 0 ? d1 / d2 :  
                                token.compareTo("+") == 0 ? d1 + d2 :  
                                                            d1 - d2;                 
              // Push result onto stack  
                stack.push( String.valueOf( result ));                                                  
            }                          
        }          

        return Double.valueOf(stack.pop());  
    }  

    public static void main(String[] args) throws Exception{  
        Scanner in = new Scanner(System.in);
        String reg = "((?<=[<=|>=|==|\\+|\\*|\\-|<|>|/|=])|(?=[<=|>=|==|\\+|\\*|\\-|<|>|/|=]))";
    while(true){
        try{
        System.out.println("Enter Your Expression");  
        //String[] input = "( 1 + 2 ) * ( 3 / 4 ) - ( 5 + 6 )".split(" ");
        String[] input =  in.nextLine() .split(reg);  
        String[] output = expToRPN(input);  

        // Build output RPN string minus the commas  
         System.out.print("Stack: ");
         for (String token : output) {  
                System.out.print("[ ");System.out.print(token + " "); System.out.print("]");
        }  
         System.out.println(" ");   
        // Feed the RPN string to RPNtoDouble to give result  
        Double result = RPNtoDouble( output );
        System.out.println("Answer= " + result);                
        }catch (NumberFormatException | EmptyStackException nfe){ 
            System.out.println("INVALID EXPRESSION"); }         
        }
    }
}

更新代码：补充：unaryToexp（）函数。我想要做的是每次发生“ - ”时，代码将其视为二进制文件，将其更改为“_”作为另一个运算符，此运算符将乘法值乘以-1（我首先想要的是添加[ - ] 1]和[*]到rpn堆栈）。这里仍有问题。

编译说：

Enter Your Expression
-5+3
Stack: [  ][ 5 ][ - ][ 3 ][ + ]
Exception in thread "main" java.lang.NumberFormatException: empty String
        at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:10 11)
        at java.lang.Double.valueOf(Double.java:504)
        at ExpressionParser.RPNtoDouble(ExpressionParser.java:160)
        at ExpressionParser.main(ExpressionParser.java:194)*

我认为它与Double d1 = Double.valueOf( stack.pop() );有关，因为它仍会弹出另外两个值，我只需要一个来解决一元运算符。有什么帮助吗？

public class ExpressionParser   
{  
    // Associativity constants for operators  
    private static final int LEFT_ASSOC  = 0;  
    private static final int RIGHT_ASSOC = 1;  

    // Operators  
    private static final Map<String, int[]> OPERATORS = new HashMap<String, int[]>();  
    static   
    {  
       // Map<"token", []{precendence, associativity}>      
        OPERATORS.put("-", new int[] { 0, LEFT_ASSOC });  
        OPERATORS.put("+", new int[] { 0, LEFT_ASSOC });  
        OPERATORS.put("*", new int[] { 5, LEFT_ASSOC });  
        OPERATORS.put("/", new int[] { 5, LEFT_ASSOC });
        OPERATORS.put("_", new int[] { 5, RIGHT_ASSOC }); 
    }  

    // Test if token is an operator  
    private static boolean isOperator(String token)   
    {
        return OPERATORS.containsKey(token);  
    }  

    // Test associativity of operator token  
    private static boolean isAssociative(String token, int type)   
    {  
        if (!isOperator(token))   
        {  
            throw new IllegalArgumentException("Invalid token: " + token);  
        }  

        if (OPERATORS.get(token)[1] == type) {  
            return true;  
        }

        return false;  
    }  

    // Compare precedence of operators.      
    private static final int cmpPrecedence(String token1, String token2)   
    {  
        if (!isOperator(token1) || !isOperator(token2))   
        {  
            throw new IllegalArgumentException("Invalid tokens: " + token1  
                    + " " + token2);  
        }  
        return OPERATORS.get(token1)[0] - OPERATORS.get(token2)[0];  
    }  


    // CONVERT UNARY OPERATORS
    public static String[] unaryToexp(String[] inputTokens)
    {
        ArrayList<String> out = new ArrayList<String>();  
        Stack<String> stack = new Stack<String>(); 
                //if token is an unary minus
        for (String token : inputTokens)   
        {
                    if( ((token == "-") && (isOperator(stack.peek()) || stack.empty()  ))){  // 
                        token = "_";
                    }
                    else if (token == "-"){
                        token = "-";
                    }
            out.add(token);
             while (!stack.empty())  
                {  
                    out.add(stack.pop());
                }       
        }

        String[] output = new String[out.size()];  
        return out.toArray(output);  
    }

    // Convert infix expression format into reverse Polish notation  
    public static String[] expToRPN(String[] inputTokens)   
    {  
        ArrayList<String> out = new ArrayList<String>();  
        Stack<String> stack = new Stack<String>();  

        // For each token  
        for (String token : inputTokens)   
        { 
            // If token is an operator  
            if (isOperator(token))   
            {  
                // While stack not empty AND stack top element   
                // is an operator 

                while (!stack.empty() && isOperator(stack.peek()))
                {                      
                    if ((isAssociative(token, LEFT_ASSOC)         && 
                         cmpPrecedence(token, stack.peek()) <= 0) ||   
                        (isAssociative(token, RIGHT_ASSOC)        &&   
                         cmpPrecedence(token, stack.peek()) < 0))       

                    {  
                        out.add(stack.pop());     
                        continue; 
                    }
                    break;  
                }

                // Push the new operator on the stack 
                stack.push(token);  
            }
            // If token is a left bracket '('  
            else if (token.equals("("))   
            {  
                stack.push(token);  //   
            }   
            // If token is a right bracket ')'  
            else if (token.equals(")"))   
            {                  
                while (!stack.empty() && !stack.peek().equals("("))   
                {  
                    out.add(stack.pop());   
                }  
                stack.pop();   
            }   
            // If token is a number  
            else   
            {  
                out.add(token);   
            }  
        }  
        while (!stack.empty())  
        {  
            out.add(stack.pop());   
        }  
        String[] output = new String[out.size()];  
        return out.toArray(output);  
    }  

   public static double RPNtoDouble(String[] tokens)  
{          
    Stack<String> stack = new Stack<String>();  

    // For each token   
    for (String token : tokens)   
    {  
        // If the token is a value push it onto the stack  
        if (!isOperator(token))   
        {  
            stack.push(token);                  
        }  
        else  
        {  
            // Token is an operator: pop top two entries  
            Double d2 = Double.valueOf( stack.pop() );  
            Double d1 = Double.valueOf( stack.pop() );  

            //Get the result  
            Double result = token.compareTo("_") == 0 ? d2 * -1 :   
                            token.compareTo("*") == 0 ? d1 * d2 :   
                            token.compareTo("/") == 0 ? d1 / d2 :  
                            token.compareTo("+") == 0 ? d1 + d2 :  
                                                        d1 - d2;    

            // Push result onto stack  
            stack.push( String.valueOf( result ));                                                  
        }                          
    }          
    return Double.valueOf(stack.pop());  
}  

    public static void main(String[] args) throws Exception{  
        Scanner in = new Scanner(System.in);
        String reg = "((?<=[<=|>=|==|\\+|\\*|\\-|\\_|<|>|/|=])|(?=[<=|>=|==|\\+|\\*|\\-|<|>|/|=]))";
    while(true){
        //try{
        System.out.println("Enter Your Expression");  
        //String[] input = "( 1 + 2 ) * ( 3 / 4 ) - ( 5 + 6 )".split(" ");
        String[] input =  in.nextLine() .split(reg); 
        String[] unary = unaryToexp(input); //.split(reg);
        String[] output = expToRPN(unary);  

        // Build output RPN string minus the commas  
         System.out.print("Stack: ");
         for (String token : output) {  
                System.out.print("[ ");System.out.print(token); System.out.print(" ]");
        }  
         System.out.println(" ");   
        // Feed the RPN string to RPNtoDouble to give result  
        Double result = RPNtoDouble( output );
        System.out.println("Answer= " + result);                
        //}catch (){ 
            //System.out.println("INVALID EXPRESSION"); }           
        }
    }   
}

Answer 1

你在这里：

private static final ScriptEngine engine = new ScriptEngineManager().getEngineByName("JavaScript");

public static String eval(String matlab_expression){
    if(matlab_expression == null){
        return "NULL";
    }
    String js_parsable_expression = matlab_expression
            .replaceAll("\\((\\-?\\d+)\\)\\^(\\-?\\d+)", "(Math.pow($1,$2))")
            .replaceAll("(\\d+)\\^(\\-?\\d+)", "Math.pow($1,$2)");
    try{
        return engine.eval(js_parsable_expression).toString();
    }catch(javax.script.ScriptException e1){
        return null; // Invalid Expression
    }
}

Answer 2

看一些示例，并尝试找到如何区分负值和运算符的规则。像：

这样的规则

 if (token is + or -) and next token is a number
 and
       (the previous token was empty
    or the prvious token was ')' or another operator)
 then it is a sign to the current value.

您可以遍历原始令牌列表，并根据此规则创建新的令牌列表。我刚刚编写了这样一个表达式求值程序，并且有一个用于标记表达式的迭代器。计划在GitHub上进行一些扩展后发布它。

编辑：这是迭代器，引用和调用应该是清楚的，由于支持变量/函数和多字符运算符，它有点复杂：

private class Tokenizer implements Iterator<String> {
    private int pos = 0;
    private String input;
    private String previousToken;

    public Tokenizer(String input) {
        this.input = input;
    }

    @Override
    public boolean hasNext() {
        return (pos < input.length());
    }

    private char peekNextChar() {
        if (pos < (input.length() - 1)) {
            return input.charAt(pos + 1);
        } else {
            return 0;
        }
    }

    @Override
    public String next() {
        StringBuilder token = new StringBuilder();
        if (pos >= input.length()) {
            return previousToken = null;
        }
        char ch = input.charAt(pos);
        while (Character.isWhitespace(ch) && pos < input.length()) {
            ch = input.charAt(++pos);
        }
        if (Character.isDigit(ch)) {
            while ((Character.isDigit(ch) || ch == decimalSeparator)
                    && (pos < input.length())) {
                token.append(input.charAt(pos++));
                ch = pos == input.length() ? 0 : input.charAt(pos);
            }
        } else if (ch == minusSign
                && Character.isDigit(peekNextChar())
                && ("(".equals(previousToken) || ",".equals(previousToken)
                        || previousToken == null || operators
                            .containsKey(previousToken))) {
            token.append(minusSign);
            pos++;
            token.append(next());
        } else if (Character.isLetter(ch)) {
            while (Character.isLetter(ch) && (pos < input.length())) {
                token.append(input.charAt(pos++));
                ch = pos == input.length() ? 0 : input.charAt(pos);
            }
        } else if (ch == '(' || ch == ')' || ch == ',') {
            token.append(ch);
            pos++;
        } else {
            while (!Character.isLetter(ch) && !Character.isDigit(ch)
                    && !Character.isWhitespace(ch) && ch != '('
                    && ch != ')' && ch != ',' && (pos < input.length())) {
                token.append(input.charAt(pos));
                pos++;
                ch = pos == input.length() ? 0 : input.charAt(pos);
                if (ch == minusSign) {
                    break;
                }
            }
            if (!operators.containsKey(token.toString())) {
                throw new ExpressionException("Unknown operator '" + token
                        + "' at position " + (pos - token.length() + 1));
            }
        }
        return previousToken = token.toString();
    }

    @Override
    public void remove() {
        throw new ExpressionException("remove() not supported");
    }

}

Answer 3

你能不能使用javascript脚本引擎？（你需要对5--2表达式进行一些调整）下面的代码输出：

3+2*1-6/3 = 3.0
3++2 = Invalid Expression
-5+2 = -3.0
5--2 = 7.0

代码：

public class Test1 {

    static ScriptEngine engine;

    public static void main(String[] args) throws Exception {
        engine = new ScriptEngineManager().getEngineByName("JavaScript");

        printValue("3+2*1-6/3");
        printValue("3++2");
        printValue("-5+2");
        printValue("5--2");
    }

    private static void printValue(String expression) {
        String adjustedExpression = expression.replaceAll("--", "- -");
        try {
            System.out.println(expression + " = " + engine.eval(adjustedExpression));
        } catch (ScriptException e) {
            System.out.println(expression + " = Invalid Expression");
        }
    }
}

Answer 4

您可以使用专门为此类任务设计的解析器生成器（如JavaCC或antlr），而不是重新发明轮子。 This is a nice example一个简单的表达式解析器和评估器，在几十行JavaCC中。

Java Expression Parser＆amp;计算器分流场算法

4 个答案: