Hello!
> Not true. The rehashing is O(n) and it has to be performed O(log n)
> times during insertion. Therefore, insertion is O(log n).
Rehashing is O(n), but the "n" is the _current_ number of items, not the
maximum one after all the insertions.
Let's assume you start with a single-entry hash table. You rehash for the
first time after inserting the first item (giving hash table of size 2),
then after the second item (=> size 4), then after the fourth item (=> size 8)
and so on. I.e., when you insert n items, the total cost of rehashing summed
over all the insertions is at most 1 + 2 + 4 + 8 + 16 + ... + 2^k (where
k=floor(log2(n))) <= 2^k+1 = O(n). That is O(1) operations per item inserted.
Have a nice fortnight
-- Martin `MJ' Mares <mj@ucw.cz> <mj@suse.cz> http://atrey.karlin.mff.cuni.cz/~mj/ MIPS: Meaningless Indicator of Processor Speed. - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
This archive was generated by hypermail 2b29 : Fri Feb 23 2001 - 21:00:25 EST