Floyd's cycle-finding algorithm

Question

I\'m trying to find this algorithm on C++ in .NET but can\'t, I found this one:

// Best solution
function boolean hasLoop(Node startNode){
  Node slowNode = Node         


        

Answer 1

The idea in the code you've found seems fine. Two fast iterators are used for convenience (although I'm positive such kind of 'convenience', like putting a lot of 'action' in the condition of while loop, should be avoided). You can rewrite it in more readable way with one variable:

while (fastNode && fastNode.next()) {
    if (fastNode.next() == slowNode || fastNode.next().next() == slowNode) {
        return true;
    }
    fastNode = fastNode.next().next();
    slowNode = slowNode.next();
}