round robin scheduling java iterators

If the list is mutable and the cost of editing it is negligible compared to I/O with the hosts, you can just rotate it:

List<String> list = Arrays.asList("one", "two", "three");
Collections.rotate(list, -1);
System.out.println(list);

You can create a new kind of Iterable that provides round-robin iteration:

public class RoundRobin<T> implements Iterable<T> {
      private List<T> coll;

      public RoundRobin(List<T> coll) { this.coll = coll; }

      public Iterator<T> iterator() { 
         return new Iterator<T>() {
            private int index = 0;

            @Override
            public boolean hasNext() {
                return true;
            }

            @Override
            public T next() {
                T res = coll.get(index);
                index = (index + 1) % coll.size();
                return res;
            }

            @Override
            public void remove() {
                throw new UnsupportedOperationException();
            }

        };
    }
}

You need to define your hosts as RoundRobin<Host>.

[FIXED based on Mirko's comment]

IMHO the standard Java API already provides an easy way to accomplish this, without resorting to external libraries or even the need to implement a custom Iterator. Simply use a Deque where you'd pull the first server, use or discard it, then append it back to the end of the Deque. Here's some sample code:

// Initialize the Deque. This might be at your class constructor. 
Deque<Host> dq = new ArrayDeque<Host>();
dq.addAll(Arrays.asList(hosts)); 

void sendJob(Job myJob) {
    boolean jobInProcess = false;
    do {
        Host host = dq.removeFirst(); // Remove the host from the top
        if(!host.isBusy()) {
            host.sendJob(myJob);
            jobInProcess = true;
        }
        dq.addLast(host); // Put the host back at the end
    } 
    while(!jobInProcess); // Might add another condition to prevent an infinite loop...    
}

This is just a sample where you always ping hosts in round robin order in a loop that only ends when one of them is available and takes the job. You could tinker with it easily to go only around the queue once (use a counter with a max set to the queue's size) or a number of times beofre throwing an exception, or sleeping in between rounds to avoid banging the hosts when all are busy.

Google collections has a utility method Iterators.cycle(Iterable<T> iterable) that does what you want.