Prolog递归计算列表中的数字 [英] Prolog recursively count numbers in a list

问题描述

我需要一个程序来计算列表中的所有数字，无论它们嵌套得多么深.我能够在数字不在另一个列表中的情况下计算数字，但是通过深度嵌套的元素递归是行不通的.到目前为止我有这个:

I need a program to count all the numbers in a list, no matter how DEEPLY NESTED they are. I was able to count numbers in the case where they were not inside another list, but recursing through deeply nested elements is not working out. I have this so far:

count([],0).
count([H|Tail], N) :-
    count(Tail, N1),
    (  number(H)
    ->N is N1 + 1
    ;   is_list(H)
    -> count(H,N)  
    ;   N = N1
    ).

所以，如果我要调用 count([a,1,[2,b],3],N)，输出应该是 N=3;但是，我只得到 N=2.有人可以帮我添加到我的第二个案例测试吗?此处所有可用的解决方案都不适用于深度嵌套的数字元素.

So, if I were to call count([a,1,[2,b],3],N), the output should be N=3; however, I only get N=2. Could someone please help me add to my second case test? All available solutions here do not work for deeply nested numerical elements.

谢谢！

推荐答案

is_list(H) 分支的代码不正确:在这种情况下，您忽略 N1，这是不正确的，您希望 N 是 N1 与 H 上的计数之和.

Your code is incorrect for the is_list(H) branch: in that case you ignore the value of N1, which is not correct, you want N to be the sum of N1 with the count on H.

完整代码:

:- use_module(library(clpfd)).

count([], 0).
count([H|T], N) :-
    count(T, N1),
    (   number(H) ->
        N #= N1 + 1
    ;   is_list(H) ->
        N #= N1 + N2,
        count(H, N2)
    ;   N1 = N
    ).

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