如何使用堆栈在一次扫描中评估中缀表达式?

时间:2012-11-16 17:23:41

标签: infix-notation stack

我想知道是否有办法使用2个堆栈在单个传递中解决中缀表达式? 堆栈可以是一个用于操作符,另一个用于操作数...

通过分流码算法求解的标准方法是将中缀表达式转换为后缀(反向抛光)然后求解。我不想先将表达式转换为后缀。

如果表达式类似2*3-(6+5)+8,如何解决?

4 个答案:

答案 0 :(得分:61)

很晚,但这是答案。

拿两个堆:

  1. operator stack {代表运营商和括号}。
  2. operand stack
  3. 算法

    如果存在要阅读的字符:

    1. 如果字符为operand推送operand stack,如果字符为(,请按operator stack
    2. 如果字符为operator
      1. 虽然operator stack的顶部的优先级不低于此字符。
      2. 来自operator的广告operator stack
      3. operands弹出两个op1op2operand stack)。
      4. op1 op op2上的operand stack存储回2.1。
    3. 如果字符为),则与2.2 - 2.4相同,直至遇到(
    4. 否则(不再需要阅读的字符):

      • 弹出式操作符直到operator stack不为空。
      • operands上的热门排名前2 push op1 op op2operand stack

      operand stack返回最高值。

答案 1 :(得分:3)

the link中给出的方法非常好。

让我引用消息来源:

We will use two stacks:

Operand stack: to keep values (numbers)  and

Operator stack: to keep operators (+, -, *, . and ^).  


In the following, “process” means, (i) pop operand stack once (value1) (ii) pop operator stack once (operator) (iii) pop operand stack again (value2) (iv) compute value1 operator  value2 (v) push the value obtained in operand stack.          


Algorithm:


Until the end of the expression is reached, get one character and perform only one of the steps (a) through (f):

(a) If the character is an operand, push it onto the operand stack.

(b) If the character is an operator, and the operator stack is empty then push it onto the operator stack.

(c) If the character is an operator and the operator stack is not empty, and the character's precedence is greater than the precedence of the stack top of operator stack, then push the character onto the operator stack.

(d) If the character is "(", then push it onto operator stack.

(e) If the character is ")", then "process" as explained above until the corresponding "(" is encountered in operator stack.  At this stage POP the operator stack and ignore "(."

(f) If cases (a), (b), (c), (d) and (e) do not apply, then process as explained above.



 When there are no more input characters, keep processing until the operator stack becomes empty.  The values left in the operand stack is the final result of the expression.

我希望这有帮助!

答案 2 :(得分:1)

  1. 创建一个空操作符堆栈。
  2. 创建一个空操作数堆栈。
  3. 表示输入字符串中的每个标记
    一个。获取中缀字符串中的下一个标记 湾如果next是一个操作数,则将其放在操作数堆栈上 C。如果下一个令牌是运营商
    • 评估运营商。
  4. 当操作符堆栈不为空时,弹出操作符和操作数(左和右),右计算左操作符并将结果推送到操作数堆栈。
  5. 来自运算符堆栈的弹出结果。

答案 3 :(得分:1)

以下是我在java中进行中缀表达式评估的尝试。如果您发现任何错误,请告诉我:))

import java.util.*;

public class ArithmeticExpressionEvaluation {

    public static void main(String[] args) {
        Scanner readExpression = new Scanner(System.in);
        System.out.print("Enter the expression: ");
        String expression = readExpression.nextLine();
        System.out.println(expression);
        System.out.println("Result: " + calculateExpression(expression));
    }

    public static long calculateExpression(String expression) {

        Stack<Long> operandStack = new Stack<>();
        Stack<Character> operatorStack = new Stack<>();

        if (!isValidExpression(expression)) {
            System.out.println("Not a valid expression to evaluate");
            return 0;
        }

        int i = 0;
        String currentInteger = null;
        while (i < expression.length()) {

            // System.out.println(expression.charAt(i));
            if (expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {

                currentInteger = expression.charAt(i) + "";
                i++;
                while (i != expression.length() && (expression.charAt(i) >= '0' && expression.charAt(i) <= '9')) {
                    currentInteger = currentInteger + expression.charAt(i);
                    i++;
                }

                operandStack.push(Long.parseLong(currentInteger));
            } else {

                if (expression.charAt(i) == ')') {

                    while (operatorStack.peek() != '(') {
                        performArithmeticOperation(operandStack, operatorStack);
                    }
                    operatorStack.pop();
                } else {

                    Character currentOperator = expression.charAt(i);
                    Character lastOperator = (operatorStack.isEmpty() ? null : operatorStack.peek());


                    if (lastOperator != null && checkPrecedence(currentOperator, lastOperator)) {
                        performArithmeticOperation(operandStack, operatorStack);
                    }
                    operatorStack.push(expression.charAt(i));

                }
                i++;
            }

        }


        while (!operatorStack.isEmpty()) {
            performArithmeticOperation(operandStack, operatorStack);
        }

    //    System.out.println(Arrays.toString(operandStack.toArray()));
    //    System.out.println(Arrays.toString(operatorStack.toArray()));

        return operandStack.pop();

    }

    public static void performArithmeticOperation(Stack<Long> operandStack, Stack<Character> operatorStack) {
        try {
            long value1 = operandStack.pop();
            long value2 = operandStack.pop();
            char operator = operatorStack.pop();

            long intermediateResult = arithmeticOperation(value1, value2, operator);
            operandStack.push(intermediateResult);
        } catch (EmptyStackException e) {
            System.out.println("Not a valid expression to evaluate");
            throw e;
        }
    }


    public static boolean checkPrecedence(Character operator1, Character operator2) {

        List<Character> precedenceList = new ArrayList<>();
        precedenceList.add('(');
        precedenceList.add(')');
        precedenceList.add('/');
        precedenceList.add('*');
        precedenceList.add('%');
        precedenceList.add('+');
        precedenceList.add('-');


        if(operator2 == '(' ){
            return false;
        }

        if (precedenceList.indexOf(operator1) > precedenceList.indexOf(operator2)) {
            return true;
        } else {
            return false;
        }

    }

    public static long arithmeticOperation(long value2, long value1, Character operator) {

        long result;

        switch (operator) {

            case '+':
                result = value1 + value2;
                break;

            case '-':
                result = value1 - value2;
                break;

            case '*':
                result = value1 * value2;
                break;

            case '/':
                result = value1 / value2;
                break;

            case '%':
                result = value1 % value2;
                break;

            default:
                result = value1 + value2;


        }
        return result;
    }


    public static boolean isValidExpression(String expression) {

        if ((!Character.isDigit(expression.charAt(0)) && !(expression.charAt(0) == '('))
                || (!Character.isDigit(expression.charAt(expression.length() - 1)) && !(expression.charAt(expression.length() - 1) == ')'))) {
            return false;
        }

        HashSet<Character> validCharactersSet = new HashSet<>();
        validCharactersSet.add('*');
        validCharactersSet.add('+');
        validCharactersSet.add('-');
        validCharactersSet.add('/');
        validCharactersSet.add('%');
        validCharactersSet.add('(');
        validCharactersSet.add(')');

        Stack<Character> validParenthesisCheck = new Stack<>();

        for (int i = 0; i < expression.length(); i++) {

            if (!Character.isDigit(expression.charAt(i)) && !validCharactersSet.contains(expression.charAt(i))) {
                return false;
            }

            if (expression.charAt(i) == '(') {
                validParenthesisCheck.push(expression.charAt(i));
            }

            if (expression.charAt(i) == ')') {

                if (validParenthesisCheck.isEmpty()) {
                    return false;
                }
                validParenthesisCheck.pop();
            }
        }

        if (validParenthesisCheck.isEmpty()) {
            return true;
        } else {
            return false;
        }
    }
}