Question

我有一个问题我在这个函数中找不到错误它有时可以很好地处理任何输入但是当输入中有括号时它会缓存{我想知道这段代码中的错误在哪里以及如何修复它，还有另一种更好的方式，而不是这种方式}

    public static String Converting_infix_expressions_to_postfix_expressions(String infix) throws Exception{
    StringTokenizer st = new StringTokenizer(infix);
    int numOF_tokens = st.countTokens();
    String postfix = "" ;
    for (int i = 1; i <= numOF_tokens; i++) {
        String term = st.nextToken();
        try {  // if it is an Float there is no problem will happen
            float x =  Float.parseFloat(term);
            postfix += x +" " ;
            System.out.println("term is number " + term);
        } catch (Exception e) {
            System.out.println("term is symbol " + term);
            if(stack.isEmpty())
                stack.push(term);
            else if(term == "(")
                stack.push(term);
            else if(term == ")"){
                while((String)stack.peek() != "(")
                    postfix += stack.pop() +" ";
                stack.pop();
            }

            else{
                int x = 0,y = 0;
                switch(term){
                case "+": x = 1; break;
                case "-": x = 1; break;
                case "*": x = 2; break;
                case "/": x = 2; break;
                }
                switch((String)stack.peek()){
                case "+": y = 1; break;
                case "-": y = 1; break;
                case "*": y = 2; break;
                case "/": y = 2; break;
                }
                if(x > y)
                    stack.push(term);
                else {
                    int x1 = x , y1 = y;
                    boolean puchedBefore = false;
                    while(x1 <= y1){
                        postfix += stack.pop() +" ";
                        if(stack.isEmpty() || stack.peek() == "(" ){
                            stack.push(term);
                            puchedBefore = true;
                            break;
                        }
                        else{
                            switch(term){
                            case "+": x1 = 1; break;
                            case "-": x1 = 1; break;
                            case "*": x1 = 2; break;
                            case "/": x1 = 2; break;
                            }
                            switch((String)stack.peek()){
                            case "+": y1 = 1; break;
                            case "-": y1 = 1; break;
                            case "*": y1 = 2; break;
                            case "/": y1 = 2; break;
                            }
                        }
                    }
                    if(!puchedBefore)
                        stack.push(term);
                }   
            }
        }
    }
    while(!stack.isEmpty()){
        postfix += stack.pop() +" ";
    }
    System.out.println("The postfix expression is : " + postfix);
    return postfix;
}

Answer 1

您的代码存在一些问题。

您应该创建自定义字符串标记生成器，因为括号和数字之间可能没有空格。例如：（5 + 6）
未正确使用try-catch块。考虑一下，首先检查字段是否为符号，然后继续将其解析为Float。这样就可以避免大部分代码出现在catch块中。
您所指的错误可通过以下第18行更改来解决。

while（！stack.isEmpty（）＆amp;＆amp;（String）stack.peek（）！=“（”）

将中缀表达式转换为后缀表达式

1 个答案: