1. Re: Subsequent storage
- Posted by Robert Craig <rds at EMAIL.MSN.COM> Jul 07, 1998
- 434 views
Robert McDougal writes: > Robert suggested using a temporary variable with this > problem. So if I set t=x[i] append something to t, and set x[i]=t > then that would take less time? I'm more confused now about this > then ever. If so, then I don't understand why that is. Wouldn't > you still be dealing with the N-squared thing when you set x[i]=t? > More confused now then ever, I meant that something like: t = x[i] for i = 1 to 10000 do t = append(t, v) end for x[i] = t will be much faster than appending to x[i] inside the loop. Regards, Rob Craig Rapid Deployment Software http://members.aol.com/FilesEu/