https://bz.apache.org/bugzilla/show_bug.cgi?id=64415
--- Comment #9 from Christopher Schultz <ch...@christopherschultz.net> --- Getting down to algorithmic complexity, the TimSort is O(n log n) -- same as merge sort. But again, we don't actually need a completely-sorted array. We just need the oldest N items. If you do a single scan of the array, that O(n) of course, so N scans of the array will be N * O(n). Note N != n. N is the number of evictions you want to do, and "n" is the number of elements in the list. So: sort+evict = O(n log n) multiple scans = N * O(n) Given that N should be relatively small, I think multiple scans wins the complexity battle. -- You are receiving this mail because: You are the assignee for the bug. --------------------------------------------------------------------- To unsubscribe, e-mail: dev-unsubscr...@tomcat.apache.org For additional commands, e-mail: dev-h...@tomcat.apache.org
