That's really no different from just sorting the list and taking the first 5. O(n log(n)). And for really large data sets, this is going to consume a lot of memory.
The method I outlined would be O(n) and doesn't force the sequence to all be resident in memory at the same time. On Fri, Sep 16, 2011 at 6:22 AM, Jim Oly <james...@gmail.com> wrote: > Using PriorityQueue should give a good, fast result. Here's my > implementation: > > (defn smallest-n [n xs] > (let [q (java.util.PriorityQueue. xs)] > (for [i (range n)] (.poll q)))) > > (smallest-n 5 (shuffle (range 100))) > ;; (0 1 2 3 4) > > Jim > > -- > You received this message because you are subscribed to the Google > Groups "Clojure" group. > To post to this group, send email to clojure@googlegroups.com > Note that posts from new members are moderated - please be patient with > your first post. > To unsubscribe from this group, send email to > clojure+unsubscr...@googlegroups.com > For more options, visit this group at > http://groups.google.com/group/clojure?hl=en > -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
