我正在制作一个转换器,它将接受中缀表达式并将它们转换为后缀表达式。
Example:
Infix: 2 * 3 - 10 / 4
Postfix: 2 3 * 10 4 / -
我有一个完全编码的方法,但它返回的后缀表达式是
2 3 * 1 0 4 / -
这样做有两个问题: 1. 主要问题是它们是 1 和 0 之间的一个空格,当它们应该在一起时(10)。2. 有很多多余的空格,输出应该和上面提供的例子一样。
我已经研究过从中缀到后缀的转换,但我无法确定如何进行比单个数字表达式转换更多的操作。
下面附上我的 postfixtoinfix 类,表达式变量以完美间距保存上例中指示的中缀。
import java.util.*;
public class InfixToPostfix
{
//Declare Instance Variables
private String expression;
private Stack<Character> stack = new Stack<Character>();
//Constructor
public InfixToPostfix(String infixExpression)
{
expression = infixExpression;
}//End of constructor
//Translate's the expression to postfix
public String translate()
{
//Declare Method Variables
String input = "";
String output = "";
char character = ' ';
char nextCharacter = ' ';
for(int x = 0; x < expression.length(); x++)
{
character = expression.charAt(x);
if(isOperator(character))
{
while(!stack.empty() && precedence(stack.peek())>= precedence(character))
output += stack.pop() + " ";
stack.push(character);
}
else if(character == '(')
{
stack.push(character);
}
else if(character == ')')
{
while(!stack.peek().equals('('))
output += stack.pop() + " ";
stack.pop();
}
else
{
if(Character.isDigit(character) && (x + 1) < expression.length() && Character.isDigit(expression.charAt(x+1)))
{
output += character;
}
else if(Character.isDigit(character))
{
output += character + " ";
}
else
{
output += character;
}
}
}//End of for
while(!stack.empty())
{
output += stack.pop() + " ";
}
return output;
}//End of translate method
//Check priority on characters
public static int precedence(char operator)
{
if(operator == '+' || operator =='-')
return 1;
else if(operator == '*' || operator == '/')
return 2;
else
return 0;
}//End of priority method
public boolean isOperator(char element)
{
if(element == '*' || element == '-' || element == '/' || element == '+')
return true;
else
return false;
}//End of isOperator method
}//End of class