我会使用一个足够快的内存映射文件（b）在$ {

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $>
MappedByteBuffer buffer = channel.map（FileChannel.MapMode.READ_ONLY，0，channel.size（））;

//完成后
channel.close（）;

假定文件小于2 GB并且需要10毫秒或更少。 p>

what's the probably fastest way of reading relatively huge files with Java's I/O-methods? My current solution uses the BufferedInputStream saving to an byte-array with 1024 bytes allocated to it. Each buffer is than saved in an ArrayList for later use. The whole process is called via a separate thread (callable-interface).

Not very fast though.

    ArrayList<byte[]> outputArr = new ArrayList<byte[]>();      
    try {
        BufferedInputStream reader = new BufferedInputStream(new FileInputStream (dir+filename));

        byte[] buffer = new byte[LIMIT]; // == 1024 
            int i = 0;
            while (reader.available() != 0) {
                reader.read(buffer);
                i++;
                if (i <= LIMIT){
                    outputArr.add(buffer);
                    i = 0;
                    buffer = null;
                    buffer = new byte[LIMIT];
                }
                else continue;              
            }

         System.out.println("FileReader-Elements: "+outputArr.size()+" w. "+buffer.length+" byte each.");   

I would use a memory mapped file which is fast enough to do in the same thread.

final FileChannel channel = new FileInputStream(fileName).getChannel();
MappedByteBuffer buffer = channel.map(FileChannel.MapMode.READ_ONLY, 0, channel.size());

// when finished
channel.close();

This assumes the file is smaller than 2 GB and will take 10 milli-seconds or less.