Java或Scala.如何将像\ x22这样的字符转换为字符串

我有一个看起来像这样的字符串:

I have a string that looks like this:

{\x22documentReferer\x22:\x22http:\x5C/\x5C/pikabu.ru\x5C/freshitems.php\x22}

如何将其转换为可读的JSON?

How could I convert this into a readable JSON?

我发现了不同的慢速解决方案，例如此处使用regEx

I've found different slow solutions like here with regEx

已经尝试过:

URL.decode
StringEscapeUtils
JSON.parse // from different libraries 

例如python有简单的解决方案，例如从'string_escape'

For example python has simple solution like decode from 'string_escape'

链接的可能重复项适用于Python，我的问题是有关Java或Scala的

Linked possible duplicate applies to Python, and my question is about Java or Scala

我正在使用的工作但也很慢的解决方案来自此处:

Working but also very slow solution I'm using now is from here:

 def unescape(oldstr: String): String = {
val newstr = new StringBuilder(oldstr.length)
var saw_backslash = false
var i = 0
while (i < oldstr.length) {
  {
    val cp = oldstr.codePointAt(i)
    if (!saw_backslash) {
      if (cp == '\\') saw_backslash = true
      else newstr.append(cp.toChar)
    } else {
      if (cp == '\\') {
        saw_backslash = false
        newstr.append('\\')
        newstr.append('\\')
      } else {
        if (cp == 'x') {
          if (i + 2 > oldstr.length) die("string too short for \\x escape")
          i += 1
          var value = 0
          try
            value = Integer.parseInt(oldstr.substring(i, i + 2), 16)
          catch {
            case nfe: NumberFormatException =>
              die("invalid hex value for \\x escape")
          }
          newstr.append(value.toChar)
          i += 1
        }
        else {
          newstr.append('\\')
          newstr.append(cp.toChar)
        }
        saw_backslash = false
      }
    }
  }
  i += 1
}
    if (saw_backslash) newstr.append('\\')
    newstr.toString
  }

private def die(msg: String) {
  throw new IllegalArgumentException(msg)
}