Clojure中的99个问题（将列表元素的连续重复项打包到子列表中）答案

【问题标题】：99 Problems number 9 in Clojure (Pack consecutive duplicates of list elements into sublists)Clojure中的99个问题（将列表元素的连续重复项打包到子列表中）
【发布时间】：2013-05-27 17:44:36
【问题描述】：

这是处理带有重复元素的单个列表的 nieve 案例，我在处理一些嵌套列表时遇到了麻烦，所以我想先写一个简单的案例。所以我有：

    (defn packDuplicatesIntoLists [listOfElements l e]
      (if(= e 'nil)
        'true
        (if(= () listOfElements)
          (if 
            (= e '())
            l
            (list l e)
          )
          (if 
              (= (first listOfElements) (first e) )
              (packDuplicatesIntoLists (rest listOfElements) l (cons (first listOfElements) e))
      (packDuplicatesIntoLists (rest listOfElements) (list l e) (first listOfElements))
          )
        )

) )

    (packDuplicatesIntoLists '(2) '(1 1) '(2 2))  (packDuplicatesIntoLists '() '(1 1) '(2 2))  (packDuplicatesIntoLists '() '() '()) (packDuplicatesIntoLists '(1 1 1 2 2 2 3 3 3 4 4 4 4) '() '())

但是 (packDuplicatesIntoLists (rest listOfElements) (list l e) (first listOfElements)) 给我添麻烦了，

    #'NintyNineProblems.LearnSpace/packDuplicatesIntoLists
    ((1 1) (2 2 2))
    ((1 1) (2 2))
    ()
    IllegalArgumentException Don't know how to create ISeq from: java.lang.Long  clojure.lang.RT.seqFrom (RT.java:505

那行有什么问题？

【问题讨论】：

哦，哎呀，那个 lise 应该是：(packDuplicatesIntoLists (rest listOfElements) (list l e) (list (first listOfElements)))，第三个 arg 假定是一个列表
所以没关系：}
(defn packDuplicatesIntoLists [listOfElements l e] (if(= '() listOfElements) (if (= e '()) l (cons e l) ) (if (= (first listOfElements) (first e ) ) (packDuplicatesIntoLists (rest listOfElements) l (cons (first listOfElements) e)) (packDuplicatesIntoLists (rest listOfElements) (con e l) (list (first listOfElements))) ) ) (packDuplicatesIntoLists '(2 2 2 4 4 4 5 5 5 8 8 8 6 9 9 9) '() '()) ((9 9 9) (6) (8 8 8) (5 5 5) (4 4 4) (2 2 2) s())

标签： clojure lisp

【解决方案1】：

所以，它仍然反转 lsit，但尾递归

     (defn packDuplicatesIntoLists [listOfElements l e]
      (if(= '() listOfElements)
       (cons e l)
       (if 
        (= (first listOfElements) (first e) )
         (recur (rest listOfElements) l (cons (first listOfElements) e))
         (if (= '() e)
          (recur (rest listOfElements) l (list (first listOfElements)))
          (recur (rest listOfElements) (cons e l) (list (first listOfElements)))
         )
        )
       )
      )

(packDuplicatesIntoLists '(2 2 2 4 4 4 5 5 5 8 8 8 6 9 9 9 9 9)'()'()) (packDuplicatesIntoLists '() '() '()) (packDuplicatesIntoLists 'nil '() '())

【讨论】：