如何转换UUID为日期格式2011-04-22？

例如，我有这样的UUID

118ffe80-466b-11e1-b5a5-5732cf729524.

如何将其转换为日期格式？

我试过了

 String uuid="118ffe80-466b-11e1-b5a5-5732cf729524"; 
    UUID uid = UUID.fromString(uuid);
    long ls=convertTime(uid.timeStamp()); // it returns long value

    public String convertTime(long time){
            System.out.println("====="+time);
            Date date = new Date(time);
            Format format = new SimpleDateFormat("yyyy/MM/dd");
            return format.format(date).toString();
        }

我得到的输出：4294744/11/02

同样的情况对于perl来说很好用

$uuid='ef802820-46b3-11e2-bf3a-47ef6b3e28e2';
$uuid =~ s/-//g;

my $timelow = hex substr( $uuid, 2 * 0,     2 * 4 );
my $timemid = hex substr( $uuid, 2 * 4,     2 * 2 );
my $version = hex substr( $uuid, 2 * 6,     1 );
my $timehi  = hex substr( $uuid, 2 * 6 + 1, 2 * 2 - 1 );

my $time = ( $timehi * ( 2**16 ) + $timemid ) * ( 2**32 ) + $timelow;
my $epoc = int( $time / 10000000 ) - 12219292800;
my $nano = $time - int( $time / 10000000 ) * 10000000;

#$time_date = scalar localtime $epoc;
#print strftime( '%d-%m-%Y %H:%M:%S', localtime($epoc) );
#print "\n Time: ", scalar localtime $epoc, " +", $nano / 10000, "ms\n";