4 Aug
2011
4 Aug
'11
5:32 p.m.
El 04/08/2011 19:29, Phil Endecott escribió:
Hi John,
John Bytheway wrote:
No; the existing complexity is better. Let S=size() and N=aS. You have
N log(S+N)
Sorry, I'm lost already. What is N log(S+N) supposed to be the complexity of?
I guess it means the number of comparisons when inserting N elements in a sorted vector/tree. N insertions, each one takes logarithmic complexity (log2(S+N) comparisons). Ion