I think I get it. With doubling, the sum of all the work done for resizing is roughly (written chonologically backwards): n/2 + n/4 + ... ~= n, and O(n) + O(n) is still O(n). Thanks!
Yep! This is generally referred to as "amortized linear time". It's the difference between considering the cost "per operation" vs. "per algorithm". The former is technically correct (as an upper bound), but too pessimistic when you consider the algorithm as a whole.
