本文介绍了在Clojure中获取一棵树的钥匙链的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想从每根到每一个假都得到一棵树的钥匙链。

I wanted to get key-chains of a tree, from every root to every leave.

例如输入树:

{"b" {:term true}, "a" {"y" {:term true}, "x" {:term true}}}

我期望的输出是:

(("b" :term) ("a" "x" :term) ("a" "y" :term)) or ["b" "ax" "ay"]

我想出了一个尾递归,它很好用:

I've figured out a tail-recursion, it worked fine:

(defn down-strings [trie]
  "this one works. 80msecs for address.txt, not bad at all"
  (lazy-seq
    (loop [strings [],trie trie]
      (if (empty? trie) strings
        (let [[k v] (first trie)]
          (recur (if (:term v) (conj strings k) strings)
                 (reduce
                   conj
                   (rest trie)
                   (map (fn [[dk dv]] [(str k dk) dv]) (dissoc v :term)))))))))

现在我遇到麻烦了与我的记录紧急练习:

Now I have trouble with my recursive practice:

(defn down-strings [trie]
  (if (map? trie)
      (map (fn [[k v]] (conj (down-strings v) k)) trie)
      []))

输出为:

(("b" [:term]) ("a" ("y" [:term]) ("x" [:term])))

我已尽一切努力,无法解决此问题。

I tried everything I could, couldn't fix this.

推荐答案

(defn down-strings [trie]
  (mapcat
    (fn [[k v]]
      (if (map? v) (map (partial str k) (down-strings v)) [""]))
    trie))

例如,

(down-strings {"b" {:term true}, "a" {"y" {:term true}, "x" {:term true}}})
;("ax" "ay" "b")

我希望它比要慢很多。

I expect this to be quite a bit slower than @noisesmith's solution.

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08-20 04:51