需要帮助将后缀转换为使用堆栈的Infix

Question

作为家庭作业的一部分，我编写了代码将后缀转换为完全括号大小的infix，但这段代码只能转换带有个位数的infix表达式。我需要帮助，以转换2或更多数字的infix表达式。

//Here's my code. My class doesn't use collection in JAVA.
//Classes and Interfaces for stack, list, and tree are provided.
private static final String DIGITS = "0123456789";

public static String convertPostfixtoInfix(String toPostfix)
{
    LinkedStack<String> s = new LinkedStack<>();

    for(int i=0; i<toPostfix.length(); i++)
    {
        if(DIGITS.indexOf(toPostfix.charAt(i)) != -1)
        {
            s.push(toPostfix.charAt(i)+"");
        }
        else if(toPostfix.charAt(i) == " ");{}//do nothing for blank.
        else
        {
            String temp = "";
            temp += toPostfix.charAt(i);

            String num1 = s.top();
            s.pop();
            String num2 = s.top();
            s.pop();
            s.push("(" + num2 + temp + num1 + ")");
        }
    }

    return s.top();//top() is same as peek() method.
}

例如，有了这些代码，

输入：4 5-9 2 1+/* 输出：(4-5)*(9/(2+1)) 输入：40 5-9 20 1+/* 输出：(9*(2/(0+1)

Answer 1

这是你怎么做的。

首先，注意一点。这一行代码是多余的：

private static final String DIGITS = "0123456789";

如果您想检查一个字符是否是一个数字，您可以简单地用

Character.isDigit();

但为了简单起见，我保留了这一行。

现在，回到你的代码。为了提供解析多个数字的功能，您所要做的就是在遇到一个数字时循环遍历输入字符串，直到第一个非数字字符。

我对您的代码做了一些修改，向您展示了它应该如何工作的基本想法：

private static final String DIGITS = "0123456789";

public static String convertPostfixtoInfix(String toPostfix)
{
    LinkedStack<String> s = new LinkedStack<>();
    StringBuilder digitBuffer = new StringBuilder();  

    /* I've changed the 'for' to 'while' loop, 
       because we have to increment i variable inside the loop, 
       which is considered as a bad practice if done inside 'for' loop
    */
    int i = 0;
    while(i < toPostfix.length()) 
    {
        if(DIGITS.indexOf(toPostfix.charAt(i)) != -1)
        {
            //when a digit is encountered, just loop through toPostfix while the first non-digit char is encountered ...
            while (DIGITS.indexOf(toPostfix.charAt(i)) != -1) {
                digitBuffer.append(toPostfix.charAt(i++)); //... and add it to the digitBuffer
            }
            s.push(digitBuffer.toString());
            digitBuffer.setLength(0); //erase the buffer
        }
        //this if-else can also be replace with only one "if (toPostfix.charAt(i) != ' ')"
        else if(toPostfix.charAt(i) == ' ');{}//do nothing for blank.
        else
        {
            String temp = "";
            temp += toPostfix.charAt(i);

            String num1 = s.top();
            s.pop();
            String num2 = s.top();
            s.pop();
            s.push("(" + num2 + temp + num1 + ")");
        }
        i++;
    }

    return s.top();//top() is same as peek() method.
}

输入：40 5-9 20 1+/* 输出：((40-5)*(9/(20+1)