List Prolog列表与递归
我在理解Prolog中的人为递归时遇到问题 一些仅分别附加到开始和结束的帮助器谓词:List Prolog列表与递归,list,recursion,prolog,List,Recursion,Prolog,我在理解Prolog中的人为递归时遇到问题 一些仅分别附加到开始和结束的帮助器谓词: add_number(Numbers, N, NewNumbers). add_letter(Letters, L, NewLetters). 我的目标是获取一个字母和数字列表,并返回两个列表:一个按外观顺序排列的数字列表,递增1;和一个字母列表,按与外观相反的顺序排列。我的理由如下: foo([], [], [], [], []). foo([X|Xs], Nums, NewNums, Letters,
add_number(Numbers, N, NewNumbers).
add_letter(Letters, L, NewLetters).
我的目标是获取一个字母和数字列表,并返回两个列表:一个按外观顺序排列的数字列表,递增1;和一个字母列表,按与外观相反的顺序排列。我的理由如下:
foo([], [], [], [], []).
foo([X|Xs], Nums, NewNums, Letters, Letters) :-
number(X),
X1 is X+1,
add_number(Nums, X1, NewNums),
foo(Xs, ???, ???, Letters, Letters).
foo([X|Xs], Nums, Nums, Letters, NewLetters) :-
letter(X),
add_letter(Letters, X, NewLetters),
foo(Xs, Nums, Nums, ???, ???).
第二个和第四个参数是累加器
那么应该这样称呼它:
realfoo(Xs, Nums, Letters) :- foo(Xs, [], Nums, [], Letters).
foo( List , Numbers , Letters ) :-
worker( List , [] , Numbers , [] , Letters ).
worker( [] , Numbers , Numbers , Letters , Letters ).
worker( [X|Xs] , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :-
digit(X),
X1 is X+1 ,
append( NumberAccumulator , [X1] , NumberAccumulator1 ) ,
worker( Xs , NumberAccumulator1 , Numbers , LetterAccumulator , Letters ).
worker( [X|Xs] , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :-
letter(X) ,
worker( Xs , NumberAccumulator , Numbers , [X|LetterAccumulator] , Letters ).
worker( [X|Xs] , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :-
not letter(X) ,
not digit(X) ,
worker( Xs , NumberAccumulator , Numbers , LetterAccumulator , Letters ).
letter( a ). letter( b ). letter( c ). ... letter( z ).
letter('A'). letter('B'). letter('C'). ... letter('Z').
digit('0'). digit('1'). digit('2'). ... digit('9').
如何编写此代码?使用累加器以相反顺序建立列表。不要使用
add_number
,否则您将得到一个二次时间算法,而您可以在线性时间内解决此问题
foo([], NumsR, Nums, Letters, Letters) :-
reverse(NumsR, Nums).
foo([X|Xs], NumsR, Nums, LettersR, Letters) :-
% the following is the Prolog syntax for if-then-else;
% you could also do this with two recursive clauses,
% but this option is faster because of first-argument indexing
(number(X) ->
X1 is X+1,
foo(Xs, [X1|NumsR], Nums, LettersR, Letters)
;
foo(Xs, NumsR, Nums, [X|LettersR], Letters)
).
foo([],Nums,Nums,字母,字母)
foo([X|Xs],Nums_1,Nums,Letters_1,Letters):-
编号(X),
X1是X+1,
添加数字(Nums_1、X1、Nums_2),
foo(Xs,Nums_2,Nums,Letters_1,Letters)
foo([X|Xs],Nums_1,Nums,Letters_1,Letters):-
字母(X),
添加字母(字母1、X、字母2),
foo(Xs,Nums_1,Nums,Letters_2,Letters)
添加数字(Nums_1、X、Nums_2):-
追加(Numbs_1[X],nums_2)
添加字母(字母1、X、字母2):-
附加(字母1,[X],字母2)。我会这样做:
realfoo(Xs, Nums, Letters) :- foo(Xs, [], Nums, [], Letters).
foo( List , Numbers , Letters ) :-
worker( List , [] , Numbers , [] , Letters ).
worker( [] , Numbers , Numbers , Letters , Letters ).
worker( [X|Xs] , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :-
digit(X),
X1 is X+1 ,
append( NumberAccumulator , [X1] , NumberAccumulator1 ) ,
worker( Xs , NumberAccumulator1 , Numbers , LetterAccumulator , Letters ).
worker( [X|Xs] , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :-
letter(X) ,
worker( Xs , NumberAccumulator , Numbers , [X|LetterAccumulator] , Letters ).
worker( [X|Xs] , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :-
not letter(X) ,
not digit(X) ,
worker( Xs , NumberAccumulator , Numbers , LetterAccumulator , Letters ).
letter( a ). letter( b ). letter( c ). ... letter( z ).
letter('A'). letter('B'). letter('C'). ... letter('Z').
digit('0'). digit('1'). digit('2'). ... digit('9').
因为这是一个学习练习,所以我不会推迟列表的反转:我会做显而易见的事情,并以相反的顺序构建列表,尽管性能受到了影响。我相信这个练习的重点是,你需要学会用两种方式建立列表