You can use Apache Commons IO to handle this and similar tasks.

The IOUtils type has a static method to read an InputStream and return a byte[].

InputStream is;
byte[] bytes = IOUtils.toByteArray(is);

Internally this creates a ByteArrayOutputStream and copies the bytes to the output, then calls toByteArray(). It handles large files by copying the bytes in blocks of 4KiB.

I have a zip file and after decoding it I get a byte array now I want to create a FileInputStream object with that byte[] object.

But you don't have a file. You have some data in memory. So a FileInputStream is inappropriate - there's no file for it to read from.

If possible, use a ByteArrayInputStream instead:

InputStream input = new ByteArrayInputStream(decodedHeaderFileZip);

Where possible, express your API in terms of InputStream, Reader etc rather than any specific implementation - that allows you to be flexible in which implementation you use. (What I mean is that where possible, make method parameters and return types InputStream rather than FileInputStream - so that callers don't need to provide the specific types.)

If you absolutely have to create a FileInputStream, you'll need to write the data to a file first.

Use the FileUtils#readFileToByteArray(File) from Apache Commons IO, and then create the ByteArrayInputStream using the ByteArrayInputStream(byte[]) constructor.

public static ByteArrayInputStream retrieveByteArrayInputStream(File file) {
    return new ByteArrayInputStream(FileUtils.readFileToByteArray(file));
}

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());