java - Balanced Parens 在 Java 中没有返回平衡

Question

我正在测试算术公式 (Ex. ((5-6)/(3+2)*34) ) 是否有平衡括号，但我检查左右括号的循环返回的结果不相等。我正在通过在出现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.");

我的右变量和左变量被初始化为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/看起来不错。

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.

Answer 1

您不想使用 break , 你想使用 continue ，但在这种情况下根本不需要。将循环更改为:

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，而不是继续下一个项目。因此，您只找到 1 个支架。

更新:switch -变体，正如 Chris 所问，您将在哪里使用 break ，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:我现在看到你的 for -loop 也是错误的。您正在使用:for (int i = 0; i < formula.size(); i++) .因为你用 pop , formula.size()每次执行循环时都会减少，而 i增加。因此你的循环结束得太早了。有两种方法可以解决这个问题。您可以使用 while - 像这样循环:

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

或者您可以更改 for -循环到这个:

int formulasize = formula.size();
for (int i = 0; i < formulasize; i++)
...