On Friday, 11 November 2016 at 13:30:17 UTC, RazvanN wrote:
I am trying to concatenate 2 ranges of the same type
(SortedRange in my case). I have tried merge, join and chain,
but the problem is that the result is not an object of the type
of the initial ranges. For example:
1. If I use chain(r1, r2), the result will be an object of type
Result which I cannot cast to my specific type (SortedRange).
2. If I use merge(r1, r2), the result will be an object of type
Merge!(typeof(r1), typeof(r)).
I know that I can use the .array property, but I think that
this iterates through all of my elements. Using
assumeSorted(chain(r1, r2).array) will return a SortedRange,
but I am not sure what the complexity for this operation is.
Is there a way to concatenate 2 ranges (SortedRange in my case)
in O(1) time?
I think chain of 2 sorted ranges will not necessarily be sorted.
For example, chain([1, 2, 10], [4, 5, 11]) -> [1, 2, 10, 4, 5,
11] which is not sorted.
You will need some kind of a merge to get a sorted range from 2
already sorted ranges.
Thanks,
SD
