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我认为这段代码应该解决带有括号的输入的问题 

1 * 2 + 3 * ( 5 + 2 ) - 3

并输出不带括号的后缀表达式，但输出为 

1.0 2.0 * 3.0 * + 5.0 2.0 + ( 3.0 - )

另一个例子：

9 - 5 * 8 + 19 - ( 8 + 10 )

将会：

9.0 5.0 8.0 * - 19.0 + - 8.0 10.0 + ( )

所以我想知道为什么这段代码不能处理这种情况。我希望以一种好的方式返回不带括号的字符串，而不删除表达式末尾的括号。

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 +" " ;
        } catch (Exception e) {
            if(stack.isEmpty())
                stack.push(term);
            else if(term == "(")
                stack.push(term);
            else if(term == ")"){
                while(!stack.isEmpty() &&  (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){ //when the operator in the peak of the stack is smaller than one read before from the infix excepression.
                        postfix += stack.pop() +" "; 
                        if(stack.isEmpty() || stack.peek() == "(" ){ // to put the term in it's right place if it samller than all the previos operators until the begining of this subset excepression "if it is between parentheses or not "
                            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;
}
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