获取Java程序以读取括号，括号和大括号

Question

我正在编写验证算术表达式的程序是否正确形成。 （例如：正确的形式"2 + (2-1)"，不正确的形式")2+(2-1"）

一旦它被验证，程序将计算结果。

目前，它可以很容易地计算括号中的任何内容。但是如果涉及到括号（例如，"2 [ 3 + (1) ]"）程序将验证表达式是否正确，但无法计算结果。

这是我关心的

void postfixExpression() { 
    stk.clear(); // Re-using the stack object 
    Scanner scan = new Scanner(expression); 
    char current; 
    // The algorithm for doing the conversion.... Follow the bullets 
    while (scan.hasNext()) { 
     String token = scan.next(); 

     if (isNumber(token)) 
     { 
      postfix = postfix + token + " "; 
     } else { 
      current = token.charAt(0); 

      if (isParentheses(current)) 
      { 
       if (stk.empty() || current == Constants.LEFT_NORMAL) { 

        // push this element on the stack; 
        stk.push(new Character(current)); 
       } else if (current == Constants.RIGHT_NORMAL) { 
        try { 

         Character ch = (Character) stk.pop(); 
         char top = ch.charValue(); 

         while (top != Constants.LEFT_NORMAL) { 
          postfix = postfix + top + " "; 
          ch = (Character) stk.pop(); 
          top = ch.charValue(); 
         } 

        } catch (EmptyStackException e) { 

        } 
       } 
      } else if (isOperator(current))// 
      { 
       if (stk.empty()) { 
        stk.push(new Character(current)); 
       } else { 
        try { 


         char top = (Character) stk.peek(); 
         boolean higher = hasHigherPrecedence(top, current); 

         while (top != Constants.LEFT_NORMAL && higher) { 
          postfix = postfix + stk.pop() + " "; 
          top = (Character) stk.peek(); 
         } 
         stk.push(new Character(current)); 
        } catch (EmptyStackException e) { 
         stk.push(new Character(current)); 
        } 
       } 
      }// Bullet # 3 ends 

     } 
    } // Outer loop ends 

    try { 
     while (!stk.empty()) // Bullet # 4 
     { 
      postfix = postfix + stk.pop() + " "; 
     } 
    } catch (EmptyStackException e) { 

    } 
} 

我创建了两个方法的代码：isBracket和isCurly。起初，我认为最合适的解决方案就是将这两种方法包含在parentheses中。像这样：（但它仍然在读的括号罚款）

if (isParentheses(current)) 
      { 
       if (stk.empty() || current == Constants.LEFT_NORMAL) { 

        // push this element on the stack; 
        stk.push(new Character(current)); 
       } else if (current == Constants.RIGHT_NORMAL) { 
        try { 

         Character ch = (Character) stk.pop(); 
         char top = ch.charValue(); 

         while (top != Constants.LEFT_NORMAL) { 
          postfix = postfix + top + " "; 
          ch = (Character) stk.pop(); 
          top = ch.charValue(); 
         } 

        } catch (EmptyStackException e) { 

        } 
       } 

      if (isCurly(current)) 
      { 
       if (stk.empty() || current == Constants.LEFT_CURLY) { 

        // push this element on the stack; 
        stk.push(new Character(current)); 
       } else if (current == Constants.RIGHT_CURLY) { 
        try { 

         Character ch = (Character) stk.pop(); 
         char top = ch.charValue(); 

         while (top != Constants.LEFT_CURLY) { 
          postfix = postfix + top + " "; 
          ch = (Character) stk.pop(); 
          top = ch.charValue(); 
         } 

        } catch (EmptyStackException e) { 


if (isBracket(current)) 
      { 
       if (stk.empty() || current == Constants.LEFT_SQUARE) { 

        // push this element on the stack; 
        stk.push(new Character(current)); 
       } else if (current == Constants.RIGHT_SQUARE) { 
        try { 

         Character ch = (Character) stk.pop(); 
         char top = ch.charValue(); 

         while (top != Constants.LEFT_SQUACRE) { 
          postfix = postfix + top + " "; 
          ch = (Character) stk.pop(); 
          top = ch.charValue(); 
         } 

        } catch (EmptyStackException e) { 

但该计划仍然不会考虑括号和大括号

我没有正确使用的方法从我个人理解，但我怎样才能恰当地使用它们？

Answer 1

下面是我想到采取不同的方法来解决这个问题的想法。

您可以将数字和括号拆分为两个不同的堆栈，从而更容易确保表达形式正确。

在代码的开始，你可以声明了两个堆栈变量：

Stack<Integer> numbers = new Stack<Integer>(); 
Stack<Character> operators = new Stack<Character>(); 

，然后相应地推动运营商和数字。

这

我创建了一个会证明这一点的实现使用两个Stack对象的非常快速的方法：

public double doCalculation(String input) throws DataFormatException { 
    if (input == null) { 
     return 0; 
    } 

    char[] characters = input.toCharArray(); 

    for (char character: characters) { 
     try { 
      // tries to push the number onto the number stack 
      numbers.push(Integer.parseInt("" + character)); 

     } catch (NumberFormatException e1) { 
      // if this is caught, this means the character is non-numerical 
      operators.push(character); 

     } 
    } 

    while (operators.size() > 0) { 
     int i = numbers.pop(); 
     int j = numbers.pop(); 
     char operator = operators.pop(); 

     switch (operator) { 
      case '+': 
       numbers.push(j + i); 
       break; 
      case '-': 
       numbers.push(j - i); 
       break; 
      case '*': 
       numbers.push(j * i); 
       break; 
      case '/': 
       numbers.push(j/i); 
       break; 
      case '^': 
       numbers.push((int)(Math.pow(j, i))); 
       break; 
      default: 
       throw new DataFormatException(); 
     } 
    } 

    return numbers.pop(); 
} 

只是为了好玩： 如果此代码被添加到catch块前的非数字字符推到堆栈，它会计算公式中的顺序操作方面：

char top; 
try { 
    top = operators.peek(); 
} catch (EmptyStackException e2) { 
    operators.push(character); 
    continue; 
} 

if (getValue(character) > getValue(top)) { 
    operators.push(character); 
    continue; 
} else { 
    try { 
     while (!(getValue(character) > getValue(operators.peek()))) { 
      char operator; 

      operator = operators.pop(); 

      int i = numbers.pop(); 
      int j = numbers.pop(); 

      switch (operator) { 
       case '+': 
        numbers.push(j + i); 
        break; 
       case '-': 
        numbers.push(j - i); 
        break; 
       case '*': 
        numbers.push(j * i); 
        break; 
       case '/': 
        numbers.push(j/i); 
        break; 
       case '^': 
        numbers.push((int)(Math.pow(j, i))); 
        break; 
       default: 
        throw new DataFormatException(); 
      } 

     } 
    } catch (EmptyStackException e3) { 
     operators.push(character); 
     continue; 
    } 

假设getValue()方法被相应地定义为：

public int getValue(char character) 
throws DataFormatException { 
    switch (character) { 
     case '+': 
      return 1; 
     case '-': 
      return 1; 
     case '*': 
      return 2; 
     case '/': 
      return 2; 
     case '^': 
      return 3; 
     default: 
      throw new DataFormatException(); 
    } 

}