How to sort ip address in ascending order

Question

Is there any method to sort this? Or do I just need to split it and use a loop to compare? Input

123.4.245.23
104.244.253.29
1.198.3.93
32.183.93.40
104.30.2         


        

Answer 1

Pad each fragment in IP to length 3 and then sort e.g. below:

    List ipList = new ArrayList();
    ipList.add("123.4.245.23");
    ipList.add("104.244.253.29");
    ipList.add("1.198.3.93");
    ipList.add("32.183.93.40");
    ipList.add("104.30.244.2");
    ipList.add("104.244.4.1");
    Collections.sort(ipList, new Comparator() {
        @Override
        public int compare(String o1, String o2) {
            String[] ips1 = o1.split("\\.");
            String updatedIp1 = String.format("%3s.%3s.%3s.%3s",
                                                  ips1[0],ips1[1],ips1[2],ips1[3]);
            String[] ips2 = o2.split("\\.");
            String updatedIp2 = String.format("%3s.%3s.%3s.%3s",
                                                  ips2[0],ips2[1],ips2[2],ips2[3]);
            return updatedIp1.compareTo(updatedIp2);
        }
    });
    //print the sorted IP
    for(String ip: ipList){
        System.out.println(ip);
    }

---

It prints:

1.198.3.93
32.183.93.40
104.30.244.2
104.244.4.1
104.244.253.29
123.4.245.23