获取Haskell中列表的所有排列 [英] Get all permutations of a list in Haskell

问题描述

我试图从头开始，而不使用标准库之外的库。下面是我的代码：

I'm trying to do this from scratch, without the use of a library outside the standard lib. Heres my code:

permutations :: [a] -> [[a]]
permutations (x:xs) = [x] : permutations' xs
    where permutations' (x:xs) = (:) <$> [x] <*> split xs
            split l = [[x] | x <- l]

问题是这只会产生一个非确定性计算的分支。理想情况下，我想要

The problem is that this only produces one fork of the non-deterministic computation. Ideally I'd want

(:) <$> [x] <*> ((:) <$> [x] <*> ((:) <$> [x] <*> ((:) <$> [x] <*> xs)))

但我找不到干净利落的方法。我想要的结果是这样的：

But I can't find a way to do this cleanly. My desired result is something like this:

permutations "abc" -> ["abc", "acb", "bac", "bca", "cab", "cba"]

我该如何做到这一点？

推荐答案

也许你应该使用现有的代码：

Maybe you should use existing code:

import Data.List
permutations [1,2,3,4]

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