From JDK 7 you can use Files.readAllBytes(Path).

Example:

Copyimport java.io.File;
import java.nio.file.Files;

File file;
// ...(file is initialised)...
byte[] fileContent = Files.readAllBytes(file.toPath());

In Java 7 you can use the readAllBytes() method of Files class. See below:

Path fileLocation = Paths.get("C:\\test_java\\file.txt");
byte[] data = Files.readAllBytes(fileLocation);

There are many other ways to do it see here and here

It is not very slow, at least there is not way to make it faster. BUT it is wrong. If file is big enough the method read() will not return all bytes from fist call. This method returns number of bytes it managed to read as return value.

The right way is to call this method in loop:

  public static void copy(InputStream input,
      OutputStream output,
      int bufferSize)
      throws IOException {
    byte[] buf = new byte[bufferSize];
    int bytesRead = input.read(buf);
    while (bytesRead != -1) {
      output.write(buf, 0, bytesRead);
      bytesRead = input.read(buf);
    }
    output.flush();
  }

call this as following:

ByteArrayOutputStream baos = new ByteArrayOutputStream();
copy(new FileInputStream(myfile), baos);
byte[] bytes = baos.toByteArray();

Something like this is implemented in a lot of packages, e.g. FileUtils.readFileToByteArray() mentioned by @Andrey Borisov (+1)

EDIT

I think that reason for slowness in your case is the fact that you create so huge array. Are you sure you really need it? Try to re-think your design. I believe that you do not have to read this file into array and can process data incrementally.