Once bytten, twice shy.
-rob
On Fri, Feb 28, 2020 at 10:17 AM Jesper Louis Andersen <
jesper.louis.ander...@gmail.com> wrote:
> The key observation is that you only look at a byte once.
>
> On Thu, Feb 27, 2020, 22:49 Amnon Baron Cohen <amno...@gmail.com> wrote:
>
>> You are right.
>> I had wrongly assumed that utf8.DecodeLastRuneInString has O(n) runtime.
>> But it has constant runtime as it reads at most 4 bytes at the end of the
>> string.
>>
>>
>> On Thursday, 27 February 2020 21:47:19 UTC, kortschak wrote:
>>>
>>> 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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>>> > .
>>>
>>>
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