infixToPostfix 算法但没有操作优先级

Question

如果我有这个字符串中缀表达式 2*4+3-15/2 并且我想输出后缀表达式而不考虑像这样的操作的优先级

2 4 * 3 + 15 - 2 /

我需要在此代码示例中进行哪些修改才能“删除”该优先级。我从 geeksforgeeks https://www.geeksforgeeks.org/stack-set-2-infix-to-postfix/ 那里获取了这段代码。我觉得改变来满足我想要的有点困难。我应该从哪里开始？谢谢。

当前代码给了我这个输出：24*3+152/-

 
    private int Prec(String ch)
    {
        switch (ch)
        {
            case "+":
            case "-":
                return 1;

            case "*":
            case "/":
                return 2;

            case "^":
                return 3;
        }
        return -1;
    }

    private  boolean isNumeric(String strNum) {
        if (strNum == null) {
            return false;
        }
        try {
            double d = Double.parseDouble(strNum);
        } catch (NumberFormatException nfe) {
            return false;
        }
        return true;
    }


    private String infixToPostfix(String infixExpression){
        // initializing empty String for result
        StringBuilder postfixExpression = new StringBuilder(new String(""));
        String[] infixExp = infixExpression.split(" ");

        // initializing empty stack
        Stack<String> stack = new Stack<>();

        for (String token : infixExp) {
            System.out.println(token+" ");
            // If the scanned character is an operand, add it to output.
            if (isNumeric(token))
                postfixExpression.append(token);

                // If the scanned character is an '(', push it to the stack.
            else if (token.equals("("))
                stack.push(token);

                //  If the scanned character is an ')', pop and output from the stack
                // until an '(' is encountered.
            else if (token.equals(")")) {
                while (!stack.isEmpty() && !stack.peek().equals("("))
                    postfixExpression.append(stack.pop());

                if (!stack.isEmpty() && !stack.peek().equals("("))
                    return "Invalid Expression"; // invalid expression
                else
                    stack.pop();
            } else // an operator is encountered
            {
                while (!stack.isEmpty() && Prec(token) <= Prec(stack.peek())) {
                    if (stack.peek().equals("("))
                        return "Invalid Expression";
                    postfixExpression.append(stack.pop());
                }
                stack.push(token);
            }

        }

        // pop all the operators from the stack
        while (!stack.isEmpty()){
            if(stack.peek().equals("("))
                return "Invalid Expression";
            postfixExpression.append(stack.pop());
        }
        System.out.println(postfixExpression);
        return postfixExpression.toString();
    }

Answer 1

看来您只需要颠倒由运算符和数字组成的每一对的顺序。你可以使用正则表达式和replaceAll:

String infix = "2*4+3-15/2";
    
String postfix = infix.replaceAll("([*+-/])([0-9]+)", " $2 $1");
    
System.out.println(postfix);

输出：

2 4 * 3 + 15 - 2 /