尝试将中缀转换为后缀会返回向后输出

时间:2019-03-15 01:58:43

标签: java postfix-notation infix-notation rpn

我正在创建一个反向波兰语表示法计算器,该计算器将中缀表达式转换为后缀表达式。但是,我的运算符向后输出。例如,我输入的中缀为“ 1 + 5 *(3 * 2)”,当我运行程序时,当输出应为“ 1 5 3 2 +”时,输出为“ +1 + * * 3 2 5” * +”,我不知道为什么。

public class RPNcalc extends Stack
{
    public static void main( String[] args)
    {
      String infix = "1+5*(3*2)";
      RPNcalc test = new RPNcalc();
      String output = test.ConvertToPostfix(infix);
      System.out.println(output);
    }


    public static String ConvertToPostfix(String infix)
    {
      Stack stack1 = new Stack();
      char ch;
      String postfix = "";

      for (int i = 0; i < infix.length(); i++)
      {
        ch = infix.charAt(i);

        if (isOperator(ch))
        {
          postfix = postfix + ch + " ";
        }
        else if (ch == '(')
        {
          stack1.push(ch);
        }
        else if (ch == ')')
        {
          while (stack1.peek() != '(')
          {
            postfix = postfix + stack1.pop() + " ";
          }
          stack1.pop();
        }
        else
        {
          while (!stack1.isEmpty() && !(stack1.peek() == '(') && (precedence(ch) <= precedence(stack1.peek())))
          {
            postfix = postfix + stack1.pop() + " ";
          }
          stack1.push(ch);
        }
      }
      while (!stack1.isEmpty())
      {
        postfix = postfix + stack1.pop();
      }
      return postfix;
    }

1 个答案:

答案 0 :(得分:1)

仅缺少一个“!”:

它必须显示为:
if( ! isOperator( ch ) ) {…}

,输出正确: 1 5 3 2 * * + 评估为31

*和+之间的间距出问题了
postfix + ' ' + stack1.pop()吗?

在这里可以找到关于调车场算法的很好描述:
https://en.wikipedia.org/wiki/Shunting-yard_algorithm#Detailed_example

顺便说一句–在Java中,函数名称应以小写字母开头:
convertToPostfix( String infix )