你可以在Clojure中将插入排序制定为一个monoid吗？

问题描述

这是Clojure中插入排序的代码：

This is the code for an insertion sort in Clojure:

(defn in-sort! [data]
  (letfn [(insert ([raw x](insert [] raw x))
          ([sorted [y & raw] x]
             (if (nil? y) (conj sorted x)
             (if (<= x y ) (concat sorted [x,y] raw)
                 (recur (conj sorted y)  raw x )))))]   
    (reduce insert [] data)))
;Usage:(in-sort! [6,8,5,9,3,2,1,4,7])
;Returns: [1 2 3 4 5 6 7 8 9]

这是：

newtype OL x = OL [x]
instance Ord x => Monoid (OL x) where
  mempty = OL []
  mappend (OL xs) (OL ys) = OL (merge xs ys) where
    merge [] ys = ys
    merge xs [] = xs
    merge xs@(x : xs') ys@(y : ys')
       | x <= y = x : merge xs' ys
       | otherwise = y : merge xs ys'

isort :: Ord x => [x] -> OL x
isort = foldMap (OL . pure)

转换为尾递归与累加器。为了多样性，这里也有一种方法来部分地模拟缺少的类型类。

Here, for reference, is another version which turns the tail recursion modulo cons into tail recursion with an accumulator. For the sake of variety, here is also one way to partially simulate the absent type-classes.

(defprotocol Monoid 
  (mempty [_] ) 
  (mappend [_ xs ys]))

(defn fold-map
  [monoid f xs]
  (reduce (partial mappend monoid) (mempty monoid) (map f xs)))

---

---

(defn- ord-mappend* 
  [[x & rx :as xs] [y & ry :as ys] a] 
  (cond
    (empty? xs) (concat a ys)
    (empty? ys) (concat a xs)
    :else (if (< x y) 
             (recur rx ys (conj a x))
             (recur xs ry (conj a y)))))

(def Ord 
  (reify Monoid 
    (mempty [_] (list))
    (mappend [_ xs ys] (ord-mappend* xs ys []))))

---

---

(defn isort [xs] (fold-map Ord list xs))

(defn is-sorted? [xs] (apply < xs))

(is-sorted? (isort (shuffle (range 10000))))
;=> true (sometime later)

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