1

我正在测试一个算术公式（例如 ((5-6)/(3+2)*34) ）是否具有平衡括号，但是我检查左右括号的循环没有返回相等。我只是在提示时在控制台中输入“（）”来测试它。

for (int i = 0; i < formula.size(); i++)
  {
    char c = formula.pop();

    if (c == ')') {
        right++;
        break;
    } else if (c == '(') {
        left++;
        break;
    } else {
        break;
    }

  }// End for loop

  //System.out.println("There are " + left + " left parens, and " + right + " right parens.");

  if (right == left)
     System.out.println("The parentheses are balanced.");
  else
     System.out.println("The parentheses are NOT balanced.");

我的左右变量初始化为0，我得到的输出是有1个右括号和0个左括号。

有什么想法吗？我写的时候听起来/看起来是对的。

更新：这是我的代码更新为使用 switch case 而不是 if else。虽然仍然得到相同的输出..

for (int i = 0; i < formula.size(); i++)
  {
    char c = formula.pop();
     switch(c)
     {
        case ')':
           right++;
           break;   //Which breaks the switch, not the for
        case '(':
           left++;
           break;   //We don't need to do anything if it's neither.

     }// End switch      
  }// End for loop

更新#2：这是我最近更改的所有主要内容：

public static void main(String[ ] args) {

  //variables
  String formulaString;
  Stack<Character> formula = new Stack<Character>();
  int right = 0;
  int left = 0;

  Scanner in = new Scanner(System.in);

  System.out.println("Welcome, enter a mathmatical formula and I will "
                    + "determine if the parentheses are balanced.\n");

  formulaString = in.next();

  for (int j = 0; j < formulaString.length(); j++) {

     formula.push(formulaString.charAt(j));

  }// End for loop

  System.out.println("Preview of the formula just entered: ");
  System.out.println(formula.display());

  System.out.println("The size of the stack is: " + formula.size());
  System.out.println("/******************************************");

  for (int i = 0; i <= formula.size(); i++)
  {
    char c = formula.pop();
    System.out.println(c);
    switch(c)
    {
       case ')':
         right++;
         break;   //Which breaks the switch, not the for
      case '(':
         left++;
         break;   //We don't need to do anything if it's neither.

     }// End switch      
  }// End for loop

  System.out.println("There are " + left + " left parens, and " + right + " right parens.");

  if (right == left)
     System.out.println("The parentheses are balanced.");
  else
     System.out.println("The parentheses are NOT balanced.");

}// End main.

我现在正在测试的输入是(()). 我得到它的输出：

Preview of the formula just entered: 
[(, (, ), )]
The size of the stack is: 4
/******************************************
)
)
(
There are 1 left parens, and 2 right parens.
The parentheses are NOT balanced.
4

2 回答 2

8

你不想使用break，你想使用continue，但在这种情况下，它根本不需要。将您的循环更改为：

for (int i = 0; i < formula.size(); i++)
{
  char c = formula.pop();

  if (c == ')')
  {
    right++;
    continue; //You don't need to add this since nothing is being done after this point
  }
  else if (c == '(')
  {
    left++;
    continue; //You don't need to add this since nothing is being done after this point
  }
  //We don't need to do anything if it's neither
}

break将退出 for 循环，而不是继续下一项。因此，您只找到 1 个支架。

更新：switch正如 Chris 所问的，您将在哪里使用 -variantbreak看起来像这样：

for (int i = 0; i < formula.size(); i++)
{
  char c = formula.pop();
  switch(c)
  {
    case '(':
      left++;
      break; //Which breaks the switch, not the for
    case ')':
      right++;
      break;
    //We don't need to do anything if it's neither. 
  }
}

更新 2：我现在看到您的for-loop 也是错误的。您正在使用：for (int i = 0; i < formula.size(); i++). 因为您使用pop,formula.size()每次执行循环时都会减少，而i会增加。因此，您的循环结束得太快了。有两种方法可以解决这个问题。您可以像这样使用while-loop：

while (formula.size() > 0)
...

或者您可以将for-loop 更改为：

int formulasize = formula.size();
for (int i = 0; i < formulasize; i++)
...
于 2013-10-22T12:04:29.377 回答
0

您放置了导致循环不完整的中断

if (c == ')') {
  right++;              
} else if (c == '(') {
  left++;               
}
于 2013-10-22T12:10:09.190 回答