Skimming through but that `case` resolves to either false or a recur
so sure you can move that case to an helper fn which returns true or
false. If false then `return` else `recur`?
On 26 June 2016 at 13:43, Botond Balázs <balazsbot...@gmail.com> wrote:
> Hi,
>
> Here is my solution to 4clojure problem #177:
>
>> Write a function that takes in a string and returns truthy if all square [
>> ] round ( ) and curly { } brackets are properly paired and legally nested,
>> or returns falsey otherwise.
>
>
> (defn valid-parens?
> [s]
> (loop [[ch & chs] s stack []]
> (if (nil? ch)
> (empty? stack)
> (case ch
> (\( \[ \{) (recur chs (conj stack ch))
> \) (if (= (peek stack) \()
> (recur chs (pop stack))
> false)
> \] (if (= (peek stack) \[)
> (recur chs (pop stack))
> false)
> \} (if (= (peek stack) \{)
> (recur chs (pop stack))
> false)
> (recur chs stack)))))
>
> This works but I don't like the repetition in the case branches. But the
> repeated code contains a recur call so I can't simply refactor it into a
> helper function. How can I achieves something like the following, but
> without consuming the stack?
>
> (defn valid-parens?
> ([s]
> (valid-parens? s []))
> ([[ch & chs] stack]
> (if (nil? ch)
> (empty? stack)
> (letfn [(match? [p]
> (if (= (peek stack) p)
> (valid-parens? chs (pop stack))
> false))]
> (case ch
> (\( \[ \{) (valid-parens? chs (conj stack ch))
> \) (match? \()
> \] (match? \[)
> \} (match? \{)
> (valid-parens? chs stack))))))
>
> Thanks!
> Botond
>
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