对文件更有效的排序

Question

I have a folder with pairs of files. Pairs share filename but have different extension (in this case, they are pairs of .txt and .png). I am looking to store them into a HashMap< File, File > by pairs. Here is what I did to find the pairs:

LinkedList<File> fileList = new LinkedList<File>(Arrays.asList(fileArray));
LinkedList<File> alreadyCompared = new LinkedList<File>();

HashMap<File, File> filePairs = new HashMap<File, File>();

for (Iterator<File> itr1 = fileList.iterator(); itr1.hasNext(); ) {

    File comparator = itr1.next();

    if (!alreadyCompared.contains(comparator)) {

        String stringComparator = comparator.getName().split("\\.")[0];
        alreadyCompared.add(comparator);

        for (Iterator<File> itr2 = fileList.iterator(); itr2.hasNext(); ) {

            File compared = itr2.next();

            if (!alreadyCompared.contains(compared)) {

                String stringCompared = compared.getName().split("\\.")[0];

                if (stringComparator.equals(stringCompared)) {

                    if (comparator.getName().endsWith("txt")) {

                        filePairs.put(comparator, compared);

                    } else {

                        filePairs.put(compared, comparator);
                    }
                }
            }
        }
    }
}

return filePairs;

Now, this takes a lot of time when I have more than 1000 files to sort, and I'd like to find a more efficient way of doing it. What other way can I sort these files?

Thanks a lot!

Answer 1

Instead of putting them in an unordered Map like HashMap , I would put the elements in a TreeMap because it sorts it's keys.

Per the linked Javadoc,

A Red-Black tree based NavigableMap implementation. The map is sorted according to the natural ordering of its keys, or by a Comparator provided at map creation time, depending on which constructor is used.

Answer 2

I made up a bit of code to illustrate my comment.

HashMap<File, File> filePairs = new HashMap<File, File>();

File dir = new File("your files directory");
File [] files = dir.listFiles(new FilenameFilter() {
    @Override
    public boolean accept(File dir, String name) {
        return name.endsWith(".txt");
    }
});

String fileName = "";

for (File txtFile : files) {
    fileName = txtfile.getName().split("\\.")[0];
    pngFile = new File(dir.getAbsolutePath() + fileName + ".png");

    if (pngFile.exists()) {
        filePairs.put(txtFile, pngFile);
    }
}