括号平衡算法

Question

我正在使用 scala 中的一个小程序来检查某些表达式是否正确地形成了关于括号的打开和关闭。这是here 的问题，但我自己的程序。

def balance(chars: List[Char]): Boolean = {
  def balanced(chars: List[Char], opened: Int, closed: Int): Boolean = {
   if (chars.isEmpty && opened == closed) true
   else if (opened < closed) false
   else {
    if (chars.head == '(') balanced(chars.tail, opened + 1, closed)
    if (chars.head == ')') balanced(chars.tail, opened, closed + 1)
    else balanced(chars.tail, opened, closed)
   }
 }
  balanced(chars, 0, 0)

}

println(balance("Just (some (random) sentence).\n(That doesn't work)".toList))

问题在于，例如，它不适用于此示例。我跟踪了程序，显然当我们从递归调用返回时出现了问题，但我无法弄清楚错误是什么。

Answer 1

在

else {
    if (chars.head == '(') balanced(chars.tail, opened + 1, closed)
    if (chars.head == ')') balanced(chars.tail, opened, closed + 1)
    else balanced(chars.tail, opened, closed)
}

当您想将两个独立的if 表达式视为单个案例表达式时，您有两个独立的表达式。如果chars.head == '(' 是true，则进行递归调用，但忽略结果并评估第二个if。这将导致采用else 分支，该分支有效地忽略了在第一个表达式中找到的(。您可以使用match，例如

chars match {
  case Nil => opened == closed
  case '('::cs => balanced(cs, opened + 1, closed)
  case ')'::cs => balanced(cs, opened, closed + 1)
  case _::cs => balanced(cs, opened, closed)
}

Answer 2

你可以试试下面的代码，可能是它在上面的例子中对你有用，如果你有“sss)fff(”也有问题，所以它在逻辑上会出错

def balance(chars: List[Char]): Boolean = {
    def balrec(chars: List[Char], accr: Int, accl: Int): Boolean = {
      chars match {
        case head :: tail =>
          if (head.equals('(')) balrec(tail, accr + 1, accl)
          else if (head.equals(')') && (accr > accl || accr == 0)) balrec(tail, accr, accl + 1)
          else balrec(tail, accr, accl)
        case Nil => if (accr == accl) true else false
      }
    }

    balrec(chars, 0, 0)
  }