删除列表中的元音 [英] Delete vowels in a list

问题描述

编写一个删除元音(String、NoVowelsString)的程序，从给定的字符串中删除所有元音.

Write a program that deletes vowels (String, NoVowelsString) that deletes all vowels from a given string.

到目前为止，我已经得到了条件 vowel(X):- member(X,[a,e,i,o,u]).然后我想到了从另一个列表中删除所有元素的那个:

So far I've got the condition vowel(X):- member(X,[a,e,i,o,u]). Then I thought of the one that deletes all the elements from the other list:

delete2([],L1,L1).
delete2([H|T],L1,L3) :-
   delete2(H,L1,R2),
   delete2(T,R2,L3).

所以有了这两个，我想我可以为那些被删除的元素设置一个条件，即它们必须是 [a,e,i,o,u] 的成员.虽然我还没有到任何地方.

So having these two I thought that I could put a condition to those elements being deleted that they have to be a member of [a,e,i,o,u]. Though I still haven't got anywhere.

推荐答案

代码如下

deleteV([H|T],R):-member(H,[a,e,i,o,u]),deleteV(T,R),!.
deleteV([H|T],[H|R]):-deleteV(T,R),!.
deleteV([],[]).

它有什么作用?首先它质疑自己?它的头部是元音是->我们忽略它.不->我们需要它.如果它找到一个空列表，则构建结果列表，并在从回溯返回时将辅音附加在前面.此代码在 SWIProlog 中进行了测试.

What it does? First it question itself?It's the head a vowel Yes->We ignore it. No->We need it. If it finds an empty list, it constructs the result list, and when returning from backtracking it appends the consonats in front. This code was tested in SWIProlog.

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