# Finding and returning a loopy path in a directed graph

I wrote a function in Scala to find out and return a loopy path in a directed graph.

One of the arguments is a graph presented in an adjacent list, and the other is a start node. It returns a pair including a loopy path by a list of nodes.

I wonder if there are more elegant ways of doing this.

  def GetACycle(start: String, maps: Map[String, List[String]]): (Boolean, List[String]) = {
def explore(node: String, visits: List[String]): (Boolean, List[String]) = {
if (visits.contains(node)) (true, (visits.+:(node)).reverse)
else {
if (maps(node).isEmpty) (false, List())
else {
val id = maps(node).indexWhere(x => explore(x, visits.+:(node))._1)
if (id.!=(-1))
explore(maps(node)(id), visits.+:(node))
else
(false, List())
}
}
}
explore(start, List())
}


I felt I had to use the indexWhere in this situation, but I suppose it would have other ways to do that.

You should use an array to check if you have already visited a node and not visits.contains(node), it would give you the answer in constant time instead of linear time.
0 -> 1, 2, ..., n

where there are n nodes and there are edges from i to j iff i<j then the node i will be explored 2^i times.