如果你有一个像123xxx这样的数字，你知道，其中一个xxx必须低于321——那么下一个palindrom是123321。

或者xxx在上面，那么3留不住了，124421得下一个。

这里有一些没有保证的代码，不是很优雅，但是中间有（多个）九的情况有点毛茸茸的（19992）：

object Palindrome extends App {

def nextPalindrome (inNumber: String): String = {
  val len = inNumber.length ()
  if (len == 1 && inNumber (0) != '9') 
    "" + (inNumber.toInt + 1) else {
    val head = inNumber.substring (0, len/2)
    val tail = inNumber.reverse.substring (0, len/2)
    val h = if (head.length > 0) BigInt (head) else BigInt (0)
    val t = if (tail.length > 0) BigInt (tail) else BigInt (0)

    if (t < h) {
      if (len % 2 == 0) head + (head.reverse)
      else inNumber.substring (0, len/2 + 1) + (head.reverse)
    } else {
     if (len % 2 == 1) {
       val s2 = inNumber.substring (0, len/2 + 1) // 4=> 4
       val h2 = BigInt (s2) + 1  // 5 
       nextPalindrome (h2 + (List.fill (len/2) ('0').mkString)) // 5 + "" 
     } else {
       val h = BigInt (head) + 1
       h.toString + (h.toString.reverse)
     }
    }
  }
}

def check (in: String, expected: String) = {
  if (nextPalindrome (in) == expected) 
    println ("ok: " + in) else 
    println (" - fail: " + nextPalindrome (in) + " != " + expected + " for: " + in)
}
//
val nums = List (("12345", "12421"), // f
  ("123456", "124421"), 
  ("54321", "54345"), 
  ("654321", "654456"), 
  ("19992", "20002"),
  ("29991", "29992"),
  ("999", "1001"),
  ("31", "33"),
  ("13", "22"),
  ("9", "11"),
  ("99", "101"),
  ("131", "141"),
  ("3", "4")
)
nums.foreach (n => check (n._1, n._2))
println (nextPalindrome ("123456678901234564579898989891254392051039410809512345667890123456457989898989125439205103941080951234566789012345645798989898912543920510394108095"))

}

我想它也可以处理一百万位数的Int。

做反转不是最好的主意。最好从字符串的开头和结尾开始，逐个元素迭代和比较。即使第一个和最后一个元素不匹配，你也在浪费时间复制整个字符串并反转它。在一百万个数字的东西上，这将是一个巨大的浪费。

对于较大的数字，这比反向快几个数量级：

def isPalindrome2(someNumber:String):Boolean = {
  val len = someNumber.length;
  for(i <- 0 until len/2) {
    if(someNumber(i) != someNumber(len-i-1)) return false; 
  }
  return true;
}

甚至可能有一种更快的方法，基于镜像字符串的前半部分。我会看看我现在是否能得到它……

更新所以这应该会在几乎恒定的时间内找到下一个回文。没有循环。我只是把它划掉了，所以我相信它可以被清理掉。

def nextPalindrome(someNumber:String):String = {
  val len = someNumber.length;
  if(len==1) return "11";
  val half = scala.math.floor(len/2).toInt;
  var firstHalf = someNumber.substring(0,half);
  var secondHalf = if(len % 2 == 1) {
    someNumber.substring(half+1,len);
  } else {
    someNumber.substring(half,len);
  }

  if(BigInt(secondHalf) > BigInt(firstHalf.reverse)) {
    if(len % 2 == 1) {
      firstHalf += someNumber.substring(half, half+1);
      firstHalf = (BigInt(firstHalf)+1).toString;
      firstHalf + firstHalf.substring(0,firstHalf.length-1).reverse
    } else {
      firstHalf = (BigInt(firstHalf)+1).toString;
      firstHalf + firstHalf.reverse;
    }
  } else {
    if(len % 2 == 1) {
      firstHalf + someNumber.substring(half,half+1) + firstHalf.reverse;
    } else {
      firstHalf + firstHalf.reverse;
    }
  }
}

这是我能实现的最通用和最清晰的解决方案：
编辑：摆脱了BigInt的，现在计算百万位数需要不到一秒钟的时间。

def incStr(num: String) = {  // helper method to increment number as String
  val idx = num.lastIndexWhere('9'!=, num.length-1)
  num.take(idx) + (num.charAt(idx)+1).toChar + "0"*(num.length-idx-1)
}

def palindromeAfter(num: String) = {
  val lengthIsOdd = num.length % 2 
  val halfLength  = num.length / 2 + lengthIsOdd
  val leftHalf  = num.take(halfLength)               // first half of number (including central digit)
  val rightHalf = num.drop(halfLength - lengthIsOdd) // second half of number (also including central digit)      

  val (newLeftHalf, newLengthIsOdd) =  // we need to calculate first half of new palindrome and whether it's length is odd or even
    if (rightHalf.compareTo(leftHalf.reverse) < 0) // simplest case - input number is like 123xxx and xxx < 321
      (leftHalf, lengthIsOdd) 
    else if (leftHalf forall ('9'==))              // special case - if number is like '999...', then next palindrome will be like '10...01' and one digit longer
      ("1" + "0" * (halfLength - lengthIsOdd), 1 - lengthIsOdd)
    else                                           // other cases - increment first half of input number before making palindrome
      (incStr(leftHalf), lengthIsOdd)

  // now we can create palindrome itself
  newLeftHalf + newLeftHalf.dropRight(newLengthIsOdd).reverse
}