ZipInputStream报java.lang.IllegalArgumentException,该怎么处理(2)

public static File zipFile(String zipName, File[] files) {
      File file = new File(zipName);

      try {
         ZipOutputStream out = new ZipOutputStream(file);

         for(int i = 0; i < files.length; i++) {
            if(files[i] == null) {
               continue;
            }

            FileInputStream in = new FileInputStream(files[i]);
            ZipEntry entry = new ZipEntry(files[i].getName().replaceAll("\\[.+\\]", ""));
            out.putNextEntry(entry);
            int len = 0;
            byte[] bytes = new byte[256];

            while((len = in.read(bytes, 0, bytes.length)) != -1) {
               out.write(bytes, 0, len);
            }

            out.closeEntry();
            in.close();
         }

         out.flush();
         out.close();
      }
      catch(Exception e) {
         e.printStackTrace();
      }

      return file;
   }