交错 Prolog 列表的元素
我是 Prolog 的新手，遇到了这个练习.问题要求定义一个谓词
I am new to Prolog and came across this practice excercise. The question asks to define a predicate
zipper([[List1,List2]], Zippered). //this is two lists within one list.
此谓词应将 List1 的元素与 List2 的元素交错.
This predicate should interleave elements of List1 with elements of List2.
例如
zipper([[1,3,5,7], [2,4,6,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8].
zipper([[1,3,5], [2,4,6,7,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8].
到目前为止，我有两个不同列表的解决方案:
So far I have a solution for two different list:
zipper ([],[],Z).
zipper([X],[],[X]).
zipper([],[Y],[Y]).
zipper([X|List1],[Y|List2],[X,Y|List]) :- zipper(List1,List2,List).
我不确定如何将这个解决方案翻译成一个列表.任何关于我可以从哪里开始的建议都会非常有帮助！
I am not sure how I can translate this solution for one list. Any suggestion on where I can start would be greatly helpful!
首先你应该把 zipper ([],[],Z). 改为 zipper ([],[],[])..然后，为了使其适用于一个列表，您可以按照评论中推荐的垫子进行操作，或者您可以稍微更改一下.所以我的版本是:
Firstly you should change zipper ([],[],Z). to zipper ([],[],[]).. Then to make it work for one list you could do what mat recommended in the comment or you could change it a little. So my version is:
zipper([],[],[]).
zipper([X,[]],X).
zipper([[],Y],Y).
zipper([[X|List1],[Y|List2]],[X,Y|List]) :- zipper([List1,List2],List).
对于您的示例:
?- zipper([[1,3,5,7], [2,4,6,8]], Zippered).
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
false.
?- zipper([[1,3,5],[2,4,6,7,8]],Zippered).
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
false.