删除列表的中间元素 | 删除列表的中间元素

本文介绍了删除列表的中间元素的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我想编写一个Prolog程序，以将奇数列表中的中间元素删除到另一个列表中.

例如，如果我们给出: delete_mid([1,2,3,4,5]，L)，那么它将产生: L = [1,2,4，5] 作为答案.

解决方案

我很惊讶也有点难过，到目前为止，这两个答案都没有采用最明显的方法；当然，您已经在学校听说过它(我怀疑这可能是OP所期望的).

然而，解释或立即做有点困难，所以首先，这里是找到中间元素的谓词:

  list_mid([H | T]，Mid):-list_mid_1(T，T，H，Mid).list_mid_1([[,, _，Mid，Mid).list_mid_1([_，_ | Fast]，[S | Slow]，__，Mid):-list_mid_1(快，慢，S，中).

我希望名字很明显.

 ?-list_mid([]，Mid).错误的.?-list_mid([x]，Mid).中= x.?-list_mid([a，x，b]，Mid).中= x.?-list_mid([a，a，x，b，b]，Mid).中= x.?-list_mid([a，a，x，b]，Mid).错误的.

似乎可以正常工作.现在，我可以尝试添加该零件，以保持其当前所扔掉的东西.

我很忙，所以花了一段时间.同时，劳布绍格的答案正是我所想到的.我没有看到它，而是这样写的:

  delete_mid([H | T]，L):-delete_mid_1(T，T，H，L).delete_mid_1([[，Rest，_，Rest).delete_mid_1([_，_ | Fast]，[H | Slow]，Prev，[Prev | Back]):-delete_mid_1(快，慢，H，后退).

这不像Raubsauger的解决方案那么整洁，但似乎其他解决方案都是相同的.对于测试用例，它以@false终止.

我认为 list_middle/2 谓词就足够了；再次让我感到惊讶和难过的是，只有劳勃绍格(Raubsauger)看到了(或已经知道).

Undtäglichgrüßtdas Murmeltier

I want to write a Prolog program to delete the middle element from an odd-numbered list into another list.

For example, If we give : delete_mid([1,2,3,4,5],L) then it will produce : L = [1,2,4,5] as answer.

解决方案

I am surprised and a bit saddened that neither answer so far takes the most obvious approach; surely you've heard about it in school (and I suspect it might be what OP is expected to do).

It is however a bit difficult to explain or do at once, so first, here is a predicate to find the middle element:

list_mid([H|T], Mid) :-
    list_mid_1(T, T, H, Mid).

list_mid_1([], _, Mid, Mid).
list_mid_1([_,_|Fast], [S|Slow], _, Mid) :-
    list_mid_1(Fast, Slow, S, Mid).

I hope the names are obvious.

?- list_mid([], Mid).
false.

?- list_mid([x], Mid).
Mid = x.

?- list_mid([a,x,b], Mid).
Mid = x.

?- list_mid([a,a,x,b,b], Mid).
Mid = x.

?- list_mid([a,a,x,b], Mid).
false.

Seems to work. Now, I can try and add the part where it keeps what it throws away at the moment.

I was busy so this took a while. In the meantime, the answer by Raubsauger is exactly what I had in mind. I did not see it and instead wrote this:

delete_mid([H|T], L) :-
    delete_mid_1(T, T, H, L).

delete_mid_1([], Rest, _, Rest).
delete_mid_1([_,_|Fast], [H|Slow], Prev, [Prev|Back]) :-
    delete_mid_1(Fast, Slow, H, Back).

It is not as neat as the solution by Raubsauger but it seems it is otherwise the same solution. It terminates for the test cases by @false.

I thought the list_middle/2 predicate was enough; I am again surprised and a bit saddened that only Raubsauger saw it (or already knew that about it).

Und täglich grüßt das Murmeltier

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