L99 Problem solution (P01 - P11)

gistfile1.clj

;;P01
;;command
(defn my-last [fseq] (list(last fseq)))

;;P02
(defn my-but-last [fseq] (list (last(butlast fseq)) (last fseq))
;;other
(take-last 2 '(a b c)) 

;;P03
(defn element-at [fseq kth] (nth fseq (- kth 1)))

;;P04 
;;これは間違い。数だけを抜き出すという課題と理解してしまった
(defn find-number [fseq] (remove false? (map #(if (number? %) % false) fseq)))

;;answer
(count '(a b c))

;;P05
(reverse '(a b c d))

;;P06
;;半分にして比較するという馬鹿みたいな行為。
(defn palindrome-check [fseq]
  (= (take (int (/ (count fseq) 2)) fseq) (reverse (take-last (int (/ (count fseq) 2)) fseq))))

;;other
;;参考：http://neuralgorithm.org/documents/l-99-in-clojure/
;;対称ならば、反対にしても一緒になるはず。
(defn palindrome-check-other [fseq]
  (= fseq (reverse fseq)))

;;P07
(flatten '(a (b (c d) e)))

;;P08
;;馬鹿みたいな分岐の嵐
(defn compress [fseq]
  (remove false? (#(loop[lseq fseq lpast nil lresult '()]
                     (if (= (next lseq) nil) 
                       (if (= lpast (first lseq)) lresult 
                                                  (flatten (cons lresult (list (first lseq)))))
                       (if (= lpast (first lseq))(recur (rest lseq) lpast lresult)
                                                 (recur (rest lseq) (first lseq) (flatten (cons lresult (list (first lseq)))))))
                     ))))
;;other
;;参考：http://neuralgorithm.org/documents/l-99-in-clojure/
;;アルゴリズムとしては次のとおり
;; '(a   a b b c d d) 　と
;; '(nil a a b b c d d) を用意。
;;もし前と後が一緒なら、mapで同じ位置の要素を取り出した場合、一緒になる。
;;違うなら、その要素を返してやればいい。

(defn p08 [fseq]
  (let [shifted-fseq (cons nil fseq)]
    (remove nil? (map #(if (not= %1 %2) %1) fseq shifted-fseq))))

;;P09
;;ifから卒業したほうがいい(ifの乱用は脳を悪くします！)
(defn pack [fseq]
  (reverse (remove nil? (#(loop[lseq fseq lpast nil lresult '() worker nil]
                     (if (= (next lseq) nil) 
                       (into lresult (list (into worker (list (first lseq)))))
                       (if (= lpast (first lseq))(recur (rest lseq) lpast lresult (into worker (list (first lseq))))
                                                 (recur (rest lseq) (first lseq) (into lresult (list worker)) (list (first lseq)) ))))
                     ))))
;;other
;;参考：http://cadr.g.hatena.ne.jp/g000001/20080926/1222382426
;;肝はletの使いどころとcond (which-caseみたいなもの)の使い方か。
;;3重くらいでまとめられそうなものは、condを使えば、見た目としてはすっきりなる
(defn pack [coll]
  (loop [coll coll, tem [], acc [] ]
    (let [[car & cdr] coll]
      (cond (empty? coll) 
            (reverse (cons tem acc))
            ;;
            (or (= car (first tem)) (empty? tem))
            (recur cdr (cons car tem) acc)
            ;;
            :else
            (recur cdr (list car) (cons tem acc))))))

;;P10
;;本当はP09を呼び出して解く問題。
(defn encode [fseq]
  (reverse (remove nil? (#(loop[lseq fseq lpast nil lresult '() worker nil]
                           (cond (empty? lseq)             (into lresult (list (list worker lpast)))
                                 (nil? worker)             (recur (rest lseq) (first lseq) lresult 1)
                                 (= lpast (first lseq))    (recur (rest lseq) lpast lresult (inc worker))
                                 (not= lpast (first lseq)) (recur (rest lseq) (first lseq) (into lresult (list (list worker lpast))) 1)
                                 ))))))

;;P11
;;せっかくなのでP10を呼び出して解答
(defn encode-modifed [fseq]
  (reverse (#(loop [lresult '() workseq (encode fseq)]
               (cond (empty? workseq)                 lresult
                     (= 1 (first (first workseq)))    (recur (into lresult (list (second (first workseq)))) (rest workseq))
                     :else                            (recur (into lresult (list (first workseq))) (rest workseq))
               )))))

;;P12
(defn decode[fseq] 
  (reverse (#(loop [lresult '() workseq fseq]
              (if (empty? workseq) lresult
                  (recur (into lresult (repeat (first(first workseq)) (second(first workseq)) )) (rest workseq) )
                )))))