獲取Java程序以讀取括號，括號和大括號

Question

我正在編寫驗證算術表達式的程序是否正確形成。 （例如：正確的形式"2 + (2-1)"，不正確的形式")2+(2-1"）

一旦它被驗證，程序將計算結果。

目前，它可以很容易地計算括號中的任何內容。但是如果涉及到括號（例如，"2 [ 3 + (1) ]"）程序將驗證表達式是否正確，但無法計算結果。

這是我關心的

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) { 

    } 
} 

我創建了兩個方法的代碼：isBracket和isCurly。起初，我認爲最合適的解決方案就是將這兩種方法包含在parentheses中。像這樣：（但它仍然在讀的括號罰款）

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) { 

但該計劃仍然不會考慮括號和大括號

我沒有正確使用的方法從我個人理解，但我怎樣才能恰當地使用它們？

Answer 1

下面是我想到採取不同的方法來解決這個問題的想法。

您可以將數字和括號拆分爲兩個不同的堆棧，從而更容易確保表達形式正確。

在代碼的開始，你可以聲明瞭兩個堆棧變量：

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

，然後相應地推動運營商和數字。

這

我創建了一個會證明這一點的實現使用兩個Stack對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(); 
} 

只是爲了好玩： 如果此代碼被添加到catch塊前的非數字字符推到堆棧，它會計算公式中的順序操作方面：

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; 
    } 

假設getValue()方法被相應地定義爲：

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(); 
    } 

}