使用Java的递归表达式求值程序

Question

I am going to write an expression evaluator which only does addition and subtraction. I have a simple algorithm to do that; but, I have some implementation problems.

I considered an expression as (it is a String)

"(" <expression1> <operator> <expression2> ")"

Here is my algorithm

String evaluate( String expression )

   if expression is digit
      return expression

   else if expression is "(" <expression1> <operator> <expression2> ")"
      cut the brackets out of it
      expression1 = evaluate( <expression1> )
      operator = <operator>
      expression2 = evaluate( <expression2> )

   if operator is +
      expression1 + expression2

   else if operator is -
      expression1 - expression2 

My problem is parsing <expression1> , <operator> and <expression2> from the expression. How can I do that?

Note: I'm not asking for a code. All I need is an idea to do that.

Thank you,

-Ali

Answer 1

My problem is parsing <expression1>, <operator> and <expression2> from the expression

Don't do that, then :) When you see an opening bracket, do your recursive call to expression. At the end of the expresssion, either you find another operator (and so you're not at the end of the expression after all), or a right-bracket, in which case you return from the evaluate.

Answer 2

Either you use a parser generator such as JavaCUP or ANTLR . Write up a BNF of your expression and generate a parser. Here is a sample grammar that would get you started:

Expression ::= Digit
            |  LeftBracket Expression Plus Expression RightBracket
            |  LeftBracket Expression Minus Expression RightBracket
            |  LeftBracket Expression RightBracket

A "hacky" way of doing it yourself would be to look for the first ) backtrack to the closest ( look at the parenthesis free expression in between, simply split on the operator symbols and evaluate.

Answer 3

Use a StringTokenizer to split your input string into parenthesis, operators and numbers, then iterate over your tokens, making a recursive call for every open-parens, and exiting your method for every close parenthesis.

I know you didn't ask for code, but this works for valid input:

public static int eval(String expr) {
    StringTokenizer st = new StringTokenizer(expr, "()+- ", true);
    return eval(st);
}

private static int eval(StringTokenizer st) {
    int result = 0;
    String tok;
    boolean addition = true;
    while ((tok = getNextToken(st)) != null) {
        if (")".equals(tok))
            return result;
        else if ("(".equals(tok))
            result = eval(st);
        else if ("+".equals(tok))
            addition = true;
        else if ("-".equals(tok))
            addition = false;
        else if (addition)
            result += Integer.parseInt(tok);
        else
            result -= Integer.parseInt(tok);
    }
    return result;
}

private static String getNextToken(StringTokenizer st) {
    while (st.hasMoreTokens()) {
        String tok = st.nextToken().trim();
        if (tok.length() > 0)
            return tok;
    }
    return null;
}

It would need better handling of invalid input, but you get the idea...

Answer 4

I would recommend changing the infix input into postfix and then evaluating it, rather than reducing the expression infix-wise. There are already well defined algorithms for this and it doesn't come with the inherent multiple-nested-parentheses parsing problems.

Take a look at the Shunting Yard Algorithm to convert to postfix/RPN then evaluate it using a stack using Postfix Operations . This is fast (O(n)) and reliable.

HTH

Answer 5

I would suggest taking an approach that more closely resembles the one described in this old but (in my opinion) relevant series of articles on compiler design. I found that the approach of using small functions/methods that parse parts of the expression to be highly effective.

This approach allows you to decompose your parsing method into many sub-methods whose names and order of execution closely follows the EBNF you might use to describe the expressions to be parsed.

Answer 6

也许为表达式和运算符创建正则表达式 ，然后使用匹配来识别和分解您的内容。