Why? There's a single correctly sized allocation made up front and then
a linear time walk along the encoded runes with truncation after each
rune.
On Thu, 2020-02-27 at 13:05 -0800, Amnon Baron Cohen wrote:
> O(n^2)
>
> On Thursday, 27 February 2020 18:53:01 UTC, rog wrote:
> > If you really just want to reverse rune-by-rune, it's pretty
> > straightforward:
> >
> > func Reverse(s string) string {
> > r := make([]byte, 0, len(s))
> > for len(s) > 0 {
> > _, n := utf8.DecodeLastRuneInString(s)
> > i := len(s) - n
> > r = append(r, s[i:]...)
> > s = s[:i]
> > }
> > return string(r)
> > }
> >
> > That will also deal correctly with invalid utf8 encoding - all the
> > bytes of the original string will be present in the result.
> > >
>
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