Java：前缀字符串的棘手排序（ArrayLists）

Question

(No networking knowledge required whatsoever. This is purely String and Lists).

Say I have a function in place, one that accepts a list of String IPv4 dotted address, and sorts them in ascending order. (Not alphabetical, true ip long format sorting). Let's call this:

public static ArrayList<String> sortListOfIpv4s(ArrayList<String> unsortedIPv4s);

This function already works correctly. Given an input:

192.168.1.1, 8.8.8.8, 4.5.6.7, 244.244.244.244, 146.144.111.6

It will output the list:

4.5.6.7, 8.8.8.8, 146.144.111.6, 192.168.1.1, 244.244.244.244

(Let's not get into a debate on whether it should modify the list in place or return a new list. It just returns a new list. Also, the function cannot be modified because of numerous reasons. )

However, my input list looks like this:

e192.168.1.1, f8.8.8.8, e4.5.6.7, f244.244.244.244, e146.144.111.6

When I remove the prefixes (only one of e or f, NOT NECESSARILY alternating) and create a clean array to pass to the sorting function, I lose the prefix information. What I would like is an output of the type:

e4.5.6.7, f8.8.8.8, e146.144.111.6, e192.168.1.1, f244.244.244.244

Basically, prior to sorting, whatever prefix was present for each element in the unsorted list, the same prefix needs to be added back to the elements in the sorted list.

Caveats:

• An IP Address can repeat in the original list, a maximum of two times
• When repeating twice, each of the two elements will have the same prefix, guaranteed
• Sorting algorithm will not remove duplicates.

A little algorithmic help please? (Remember, we already have a function that can sort clean IPv4 String arraylists).

Answer 1

Don't remove the prefixes prior to passing it to the sorting function. Instead, in the sortListOfIpv4s method, always compare Strings using s.substring(1) , which will give you the entire string without the prefix, and add s to the resulting sorted array.

If sortListOfIpv4s is a black box and you are required to pass the prefix-free Strings, then you could cache the prefixes beforehand in a Map from prefix-free IP -> prefix :

Map<String, String> prefixMap = new HashMap<String, String>();
for (String ip : unsortedIPv4s) {
  prefixMap.put(ip.substring(1), ip.substring(0, 1));
}

Then sort and recover the prefixes from the Map :

List<String> sortedIPV4s = sortListOfIpv4s(unsortedIPv4s);
for (String ip : sortedIPV4s) {
  String prefix = prefixMap.get(ip);
  String originalIp = prefix + ip;
}

Answer 2

您的方法可以将任何前缀移动到String的末尾，对列表进行排序，然后再次遍历字符串并将前缀从末尾移回到开头。

Answer 3

You could implement Comparator:

public class IpComparator implements Comparator<String> {
    @Override
    public int compare(String ipA, String ipB) {
        return doComparison( ipA.substring(1), ipB.substring(1) );
    }
}

Then you can use it:

return Collections.sort(unsortedIPv4s, new IpComparator());