[英]Shunting-yard algorithm parsing function agruments
我試圖讓我的shunting-yard算法的實現工作。 它適用於數字和運算符。 但是當我嘗試向輸入添加函數時會出現問題。 因為函數參數在應該向右輸出時輸出到函數的左側。
測試程序
public class FrontTest
{
public static void main(String[] args)
{
String str = "cbrt ( 8 )";
System.out.println(ShuntTest.infixToPostfix(str));
}
}
算法
import java.util.*;
public class ShuntTest
{
public static String infixToPostfix(String infixStr)
{
Stack<String> operators = new Stack<String>();
Queue<String> output = new LinkedList<String>();
String[] tokens = infixStr.split("[\\s]");
StringBuilder postfixStr = new StringBuilder();
int tokensRemaining = tokens.length;
final String PAREN_LEFT = "[\\(]";
final String PAREN_RIGHT = "[\\)]";
final String FUNCTION_ARGSEP = "[\\;]";
for (int i = 0; i < tokens.length; i++)
{
if (isNumber(tokens[i]))
{
output.offer(tokens[i]);
}
else if (isFunction(tokens[i]))
{
operators.push(tokens[i]);
}
else if (tokens[i].matches(FUNCTION_ARGSEP))
{
while (!operators.empty() && operators.peek().matches(PAREN_RIGHT))
{
output.offer(operators.pop());
if (operators.empty() && !operators.peek().matches(PAREN_RIGHT))
{
throw new RuntimeException("Mismatched Parentheses.");
}
}
}
else if (isOperator(tokens[i]))
{
while (!operators.empty() && ((isOperator(operators.peek())
&& ((isLeftAssociative(tokens[i]) == true && ((operatorPrecedence(tokens[i]) <= operatorPrecedence(operators.peek()))
|| ((isLeftAssociative(tokens[i]) == false && ((operatorPrecedence(tokens[i]) < operatorPrecedence(operators.peek())))))))))))
{
output.offer(operators.pop());
}
operators.push(tokens[i]);
}
else if (!operators.empty() && tokens[i].matches(PAREN_LEFT))
{
operators.push(tokens[i]);
}
else if (!operators.empty() && tokens[i].matches(PAREN_RIGHT))
{
while (!operators.empty() && !operators.peek().matches(PAREN_LEFT))
{
output.offer(operators.pop());
}
if (!operators.empty())
{
operators.pop();
}
else if (!operators.empty() && isFunction(operators.peek()))
{
output.offer(operators.pop());
}
else if (operators.empty())
{
throw new RuntimeException("Mismatched Parentheses.");
}
}
tokensRemaining--;
}
if (tokensRemaining == 0)
{
while (!operators.empty())
{
if (operators.peek().matches(PAREN_LEFT)
|| operators.peek().matches(PAREN_RIGHT))
{
throw new RuntimeException("Mismatched Parentheses.");
}
output.offer(operators.pop());
}
}
while (!output.isEmpty())
{
while (output.size() > 1)
{
postfixStr.append(output.poll() + " ");
}
postfixStr.append(output.poll());
}
return postfixStr.toString();
}
public static boolean isNumber(String str)
{
final String NUMBER = "^0?-?\\+?\\d+(\\.\\d+)?$";
return str.matches(NUMBER) ? true : false;
}
public static boolean isOperator(String str)
{
switch (str)
{
case "^":
case "/":
case "*":
case "+":
case "-":
return true;
default:
return false;
}
}
private static boolean isLeftAssociative(String str)
{
switch (str)
{
case "^":
return false;
case "/":
case "*":
case "+":
case "-":
return true;
default:
throw new IllegalArgumentException("Operator unknown: " + str);
}
}
private static int operatorPrecedence(String str)
{
switch (str)
{
case "^":
return 4;
case "/":
case "*":
return 3;
case "+":
case "-":
return 2;
default:
throw new IllegalArgumentException("Operator unknown: " + str);
}
}
public static boolean isFunction(String str)
{
switch (str)
{
case "sin":
case "cos":
case "tan":
case "sqrt":
case "cbrt":
case "root_of":
return true;
default:
return false;
}
}
}
例1:
輸入: cbrt ( 8 )
輸出: 8 cbrt
輸出應為: cbrt 8
例2:
輸入: cbrt ( 8 ) + sqrt ( 9 )
輸出: 8 9 sqrt + cbrt
輸出應為: cbrt 8 sqrt 9 +
例3:
輸入: 5 + 4 - 9 / 2 ^ 6 * cbrt ( 8 )
輸出: 5 4 + 9 2 6 ^ / 8 cbrt * -
輸出應為: 5 4 + 9 2 6 ^ / cbrt 8 * -
想一想。 分流碼算法將中綴轉換為后綴。 不過這個:
cbrt ( 8 )
不是中綴表達式。 它是前綴表達式。
作為后綴表達式,它看起來像這樣:
8 cbrt
它作為一個中綴表達式可能看起來像留給讀者的練習,但它不是你開始的。
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