Comparing version numbers

Question

Some time ago, I read that comparing version numbers can be done using the following code snippet:

NSString *vesrion_1 = @\"1.2.1\";
NSString *version_2 = @\         


        

Answer 1

In Java you could use the following code snippet to compare versions. If one version number is shorter then the other we normalize it with zeros.

import java.util.Arrays;
import java.util.Comparator;

public class FreePlay {

    static String[] versions = {"1.5.3.2", "2.3.4", "1.0.3.4", "10.10.1.1", "1.0.2.5", "2.3.4"};
    /**
     * @param args
     */
    public static void main(String[] args) {
        System.out.println("Unsorted versions");
        for (String s: versions) {
            System.out.print("'" + s + "', ");
        }

        System.out.println("\nSorted versions");

        Arrays.sort(versions, new Comparator() {

            @Override
            public int compare(String o1, String o2) {
                String[] firstVersions = o1.split("\\.");
                String[] secondVersions = o2.split("\\.");
                int length = Math.max(firstVersions.length, secondVersions.length);
                for(int i = 0; i < length; i++) {
                    // normalize the length. If one part is short, we normalize it with zero
                    int firstVersion = i < firstVersions.length ? Integer.parseInt(firstVersions[i]) : 0;
                    int secondVersion = i < secondVersions.length ? Integer.parseInt(secondVersions[i]) : 0;
                    if(firstVersion < secondVersion)
                        return -1;
                    if(firstVersion > secondVersion)
                        return 1;
                }
                return 0;
            }});

        for (String s: versions) {
            System.out.print("'" + s + "', ");
        }
    }

}