clojure - 列出列表的所有排列 [英] clojure - list all permutations of a list

问题描述

假设我有一个这样的集合:

Say I have a set like this:

#{"word1" "word2" "word3"}

我怎样才能列出这些单词的所有排序方式，即

How could I list all ways that these words might be ordered, i.e.

word1 word2 word3
word2 word3 word1
word3 word2 word1

等

推荐答案

最简单的方法是使用 math.组合学:

The easiest way is using math.combinatorics:

user> (require '[clojure.math.combinatorics :as combo])
nil
user> (combo/permutations #{"word1" "word2" "word3"})
(("word1" "word2" "word3") ("word1" "word3" "word2") ("word2" "word1" "word3") ("word2" "word3"    "word1") ("word3" "word1" "word2") ("word3" "word2" "word1"))

我没有看过 math.combinatorics 的实现，但这是一个懒惰的版本，因为 OP 要求遵循一些代码.

I haven't looked at the math.combinatorics implementation, but here's a lazy version because OP asked for some code to follow.

(defn permutations [s]
  (lazy-seq
   (if (seq (rest s))
     (apply concat (for [x s]
                     (map #(cons x %) (permutations (remove #{x} s)))))
     [s])))

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