On Fri, Apr 21, 2017 at 12:43 AM Ian Davis <m...@iandavis.com> wrote:
>
>
>
> On Fri, 21 Apr 2017, at 03:31 AM, Ivan Kurnosov wrote:
>
> @Rob,
>
> honestly to me they look the same:
>
>
> func IsSorted(data Interface) bool {
> n := data.Len()
> for i := n - 1; i > 0; i-- {
> if data.Less(i, i-1) {
> return false
> }
> }
> return true
> }
>
>
> func IsSortedForward(data sort.Interface) bool {
> n := data.Len()
> for i := 1; i < n; i++ {
> if data.Less(i, i-1) {
> return false
> }
> }
> return true
> }
>
>
> https://play.golang.org/p/KSwwMk67ew
>
>
> How well does this handle slices of length 0 and 1?
>
It's easy to see that if n <= 1, both solutions run the loop 0 times and
return true (except if n == MinInt; see below).
The counting down solution would break (run a very long loop) if n were
uint(0), while the counting up version would not. It's not a problem though
because Len() returns int. The counting down solution does run a very long
loop if data.Len() returns MinInt, while the counting up solution returns
true immediately in that case. For all other negative values of data.Len()
both return true immediately.
The MinInt problem could be eliminated while still counting down with
for i := n; i > 1; i--
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