在 Java 中获取最新（降序上次修改）n 个文件的最佳优化方法是什么 - 无需加载大目录的所有文件

Question

Aim is to get latest 100 files. Currently it is done by scanning all files - preparing a files list - and then apply sort+limit.

this is very slow - in cases when directory is too large. So is there any way or API available which does this without loading full file list.

Currently following three approaches do not give satisfactory performance when files are in range of few thousands.

• Files.listFiles - Java 1.2
• DirectoryStream - Java 1.7
• Files.Walk - Java 1.8

Answer 1

You have to look at the attributes of each file to find its age, and you have to look at them all to find the N newest.

Your only freedom of choice is in how you do the looking. There's no need to read the file contents, for example.

I'd consider using Files.find(). This appears from its documentation to do the minimum work required.

You don't need to save all files. Track the oldest of the newest 100 seen. If the 'next' file is older than that, you don't need to keep it. Otherwise you have to figure out which of the 100 to discard. This trades off overhead of keeping an entire list for overhead of deciding what to discard. It could work in your favour if the number of files is much larger than 100.

To some extent the overhead is file-system dependent. If the last-modified time is stored in the directory entry then there's no need to look at the inode to get it. That's not under your control, of course.