检查列表是否是回文.如果没有,插入元素使其成为回文.(序言)

Question

我写了下面的代码来检查它是否是回文.我还创建了在列表不是回文时插入元素的逻辑

reverse_list(Inputlist, Outputlist) :-
   reverse(Inputlist, [], Outputlist).    

reverse([], Outputlist, Outputlist).    
reverse([Head|Tail], List1, List2) :-
   reverse(Tail, [Head|List1], List2).

printList([]).
printList([X|List]) :-
   write(X),
   write(' '),
   printList(List).

palindrome(List1) :-
   reverse_list(List1, List2),
   compareLists(List1, List1, List2, List2).

compareLists(L1, [], [], L2) :-
   write("\nList is Palindrome").    
compareLists(L1, [X|List1], [X|List2], L2) :-
   compareLists(L1, List1, List2, L2),
   !.        
compareLists(L1, [X|List1], [Y|List2], [Z|L2]) :-
   write("\nList is not Palindrome. "),
   append(L1, L2, L),
   printList(L).

代码给出了正确的输出

palindrome([a,b,c,a]).
List is not Palindrome. a b c a c b a 

palindrome([a,b,c]).
List is not Palindrome. a b c b a 

但是,对于诸如此类的输入

palindrome([a,b,c,b]).
List is not Palindrome. a b c b c b a 

然而,最佳解决方案应该是

 a b c b a

我应该加入哪些变化才能实现这一目标？

Answer 1

我认为你需要一个带有两个参数的谓词， In 和 Out ：

pal([], []).
pal([X], [X]).
pal(In, Out) :-
    % first we check if the first and last letter are the same
    (   append([H|T], [H], In)
        % we must check that the middle is a palindrome
    ->  pal(T, T1),
        append([H|T1], [H], Out)
    ;   % if not, we remove the first letter
        % and we work with the rest
        In = [H|T],
        % we compute the palindrome from T
        pal(T,T1),
        % and we complete the palindrome to
        % fit the first letter of the input
        append([H|T1], [H], Out)).

EDIT1 这段代码看起来不错，但有一个错误

? pal([a,b,c,a], P).
P = [a, b, c, b, a] .

应该是 [a,b,c,a,c,b,a] 我会尝试修复它。

EDIT2 看起来正确：

build_pal([H|T], Out):-
    pal(T,T1),
    append([H|T1], [H], Out).


pal([], []).
pal([X], [X]).
pal(In, Out) :-
    (   append([H|T], [H], In)
    ->  pal(T, T1),
        (   T = T1
        ->  append([H|T1], [H], Out)
        ;   build_pal(In, Out))
    ;   build_pal(In, Out)).

输出：

 ?- pal([a,b,c], P).
P = [a, b, c, b, a] .

 ?- pal([a,b,a], P).
P = [a, b, a] .

 ?- pal([a,b,c,b], P).
P = [a, b, c, b, a] .

 ?- pal([a,b,c,a], P).
P = [a, b, c, a, c, b, a] .

 ?- pal([a,b,a,c,a], P).
P = [a, b, a, c, a, b, a] .