IP与十进制相互转化

def ip2Long(ip: String): Long = {

val fragments = ip.split("[.]")
  var ipNum = 0L
  for (i <- 0 until fragments.length){
    ipNum =  fragments(i).toLong | ipNum << 8L
  }
  ipNum
}

二分法查找：

def binarySearch(arr: Array[(String, String, String, String)], ip: Long): Int =

{     var l = 0

var h = arr.length - 1

while (l <= h) {

var m = (l + h) / 2

if ((ip >= arr(m)._1.toLong) && (ip <= arr(m)._2.toLong))

{

return m

} else if (ip < arr(m)._1.toLong)

{

h = m - 1

}

else

{

l = m + 1

}

}

-1

}