两个列表的排列 [英] Permutations of two lists

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本文介绍了两个列表的排列的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我在OCAML中有一个char list.我想为chartruefalse的每种组合创建一个( (char * bool) list) list.

I have a char list in OCAML. I would like to create a ( (char * bool) list) list of every combination of the chars with true and false.

我猜想我必须做的是List.fold_left之类的东西,但是我不确定如何实现它.

What I have guess I have to do is something like a List.fold_left, but I am not quite sure how to pull it off.

这是我尝试过的大纲(OCAML语法,但不可运行):

This is the outline that I tried (OCAML syntax, but not run-able):

let rec var_perm var_list options = 
    match var_list with
        | [] -> options
        | x :: v' ->
            ((x, true) :: (var_perm_intern v')) :: ((x, false) :: (var_perm_intern v'))
;;

let all_options = var_perm ['a';'b'] [];;

应该返回

[
    [('a',true);('b',true)];
    [('a',true);('b',false)];
    [('a',false);('b',true)];
    [('a',false);('b'false)];
]

另一个示例:

let all_options = var_perm ['u';'w';'y'] [];;

应该退货(顺序并不重要)

should return (order is not important)

[
    [('u',false);('w',false);('y',false)];
    [('u',false);('w',false);('y',true )];
    [('u',false);('w',true );('y',false)];
    [('u',false);('w',true );('y',true )];
    [('u',true );('w',false);('y',false)];
    [('u',true );('w',false);('y',true )];
    [('u',true );('w',true );('y',false)];
    [('u',true );('w',true );('y',true )];
]

推荐答案

您已接近正确的解决方案.具体来说:

You're close to a correct solution. Specifically:

  • 您必须在递归调用中删除_intern后缀
  • "options"参数没有用(看看您如何递归调用,仅传递一个参数v'),因此您必须找出在[]情况下返回的内容
  • "v'的结果,加上true代表头部var"和"v'的结果,加上false代表头部var"的串联应写为foo @ bar而不是foo :: bar,因为这是您要串联的两个列表,而不是一个元素添加到列表中.
  • you must remove the _intern suffix in your recursive call
  • the "options" parameter is useless (look at how you do your recursive call, passing only one parameter v'), so you must find out what to return in the [] case
  • the concatenation of "the results of v', plus true for the head var" and " the results of v', plus false for the head var" should be written foo @ bar rather than foo :: bar, because those are two lists you're concatenating, not one element added to a list.

这篇关于两个列表的排列的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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