How to precalculate valid number of combinations instead of using while loop?

Question

Given List of datacenters which are dc1, dc2, dc3 and list of machines, h1, h2, h3, h4 as mentioned below -

Datacenters = dc1, dc2, dc3
Machines = h1, h2, h3, h         


        

Answer 1

Not really sure what you've asked here, but i think i can see where the problem is, each time get call get mapping you only generate 1 row. so i've rewritten the code so it generates all of them and gives you back a list of the maps. so you can do what you need with them.

public class DataCenterMapping {

    public static void main(String[] args) {

        DatacenterMachineMapping dcm = new DatacenterMachineMapping(
                Arrays.asList("dc1", "dc2", "dc3"), Arrays.asList("h1", "h2",
                        "h3", "h4"));


            List> coloHost = dcm
                    .getDatacenterMachineMappings();

            System.out.println(coloHost);

    }
}

class DatacenterMachineMapping {

    private boolean firstCall = true;
    private int hostListIndex = 0;
    private List datacenterList, hostList;

    public DatacenterMachineMapping(List datacenterList,
            List hostList) {
        this.datacenterList = datacenterList;
        this.hostList = hostList;
    }

    public List> getDatacenterMachineMappings() {
        List> grid = new ArrayList>();
        for (int i = 0; i < datacenterList.size(); i++) {

            Map datacenterMachineMapping = new HashMap();
            String[] line = new String[hostList.size()];
            for (int j = 0; j < line.length; j++) {
                int off = j + i;
                if (off >= datacenterList.size()) {
                    off -= datacenterList.size();
                }
                datacenterMachineMapping.put(hostList.get(j) ,datacenterList.get(off));
            }

            grid.add(datacenterMachineMapping);
        }
        return grid;
    }

}

example output:

[{h4=dc1, h1=dc1, h3=dc3, h2=dc2}, {h4=dc2, h1=dc2, h3=dc1, h2=dc3}, {h4=dc3, h1=dc3, h3=dc2, h2=dc1}]