Имея некоторый конечный список, вернуть список всех его префиксов, включая пустой список, в порядке возрастания их длины.

(В основном реализация функции Haskell inits.)

Детали

• Список ввода содержит числа (или другой тип, если это более удобно).
• Вывод должен быть списком списков .
• Представление может, но не обязательно должно быть функцией, любой ввод / вывод по умолчанию может быть использован.
• Есть ответ CW для всех тривиальных решений .

пример

[] -> [[]]
[42] -> [[],[42]]
[1,2,3,4] -> [[], [1], [1,2], [1,2,3], [1,2,3,4]]
[4,3,2,1] -> [[], [4], [4,3], [4,3,2], [4,3,2,1]]

Haskell , 20 байтов

Изменить: еще байта короче с совершенно другим сканированием.

Анонимная функция немного превосходит тривиальный импорт.

scanr(\_->init)=<<id

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• Используется =<<для сокращения (scanr(\_->init)=<<id) l = scanr(\_->init) l l.
• Сканирует список lсправа налево, собирая промежуточные результаты с помощью функции \_->init.
• Эта функция игнорирует просканированные элементы (они используются только для получения правильной общей длины для собранных результатов), поэтому действительно повторяет применение initк начальному значению сканирования, которое также l.

Брейкфук , 21 12 байт

-9 байт благодаря Арно, предложившему разделитель ÿвместо новых строк

-[[<]>[.>],]

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Принимает байты через STDIN без нулевых байтов и печатает серию префиксов, разделенных ÿсимволом с начальным ÿсимволом. Например, для вводаPrefixes , выход ÿÿPÿPrÿPreÿPrefÿPrefiÿPrefixÿPrefixeÿPrefixes.

Для удобства чтения, вот версия с переводом строки .

Объяснение:

-              Create a ÿ character in cell 0
 [        ,]   While input, starting with the ÿ
  [<]>           Go to the start of the string
      [.>]       Print the string
          ,      Append the input to the end of the string

JavaScript (ES6), 33 байта

a=>[b=[],...a.map(n=>b=[...b,n])]

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Как?

+--- a = input array
|
|       +--- initialize b to an empty array and include it as the first entry
|       |    of the output (whatever the input is)
|       |
|       |          +--- for each value n in a[]:
|       |          |
|       |          |        +--- append n to b[] and include this new array in
|       |          |        |    the final output
|       |          |        |
a => [b = [], ...a.map(n => b = [...b, n])]
               |                  |
               +---------+--------+
                         |
      spread syntax: expands all elements of
      the child array within the parent array

CW для всех тривиальных записей

чистый , 19 байт

Версия на Haskell работает и в Clean.

import StdLib
inits

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Haskell , 22 байта

import Data.List
inits

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Пролог (SWI) , 6 байт

prefix

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Желе , 3 байта

ṭṖƤ

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Как это работает

ṭṖƤ  Main link. Argument: A

  Ƥ  Map the link to the left over all non-empty(!) prefixes of A.
 Ṗ       Pop; remove the last element.
ṭ    Tack; append A to the resulting list.

Japt, 4 bytes

²£¯Y

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Explanation:

²       :Add an arbitrary extra item to the end of the array
 £      :For each item in the new array:
  ¯Y    : Get an array of the items that are before it

Perl 6, 13 bytes

{(),|[\,] @_}

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To explain:

In Perl 6 you can wrap an operator in square brackets as an alternate way to write a list reduction. [+] @array returns the sum of the elements in @array, [*] @array returns the product, etc. You can also precede the operator with a backslash to make a "triangular" reduction, which some languages call "scan." So [\+] @array returns a list consisting of the first element of @array, then the sum of the first two elements, then the sum of the first three elements, etc.

Here [\,] @_ is a triangular reduction over the input array @_ using the list construction operator ,. So it evaluates to a lists of lists: the first element of @_, the first two elements of @_, etc. That's almost what's needed, but the problem calls for a single empty list first. So the first element of the return list is a literal empty list (),, then the reduction over the input list is flattened into the rest of the return list with |.

Python 2, 32 bytes

f=lambda l:(l and f(l[:-1]))+[l]

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R, 40 39 bytes

function(L)lapply(0:length(L),head,x=L)

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-1 byte thanks to digEmAll

The output of R's list type is a bit weird; it uses sequential indexing, so for instance, the output for

list(1,2) is

[[1]]                     # first list element
list()

[[2]]                     # second list element
[[2]][[1]]                # first element of second list element
[1] 1


[[3]]                     # third list element
[[3]][[1]]                # first element of third list element
[1] 1

[[3]][[2]]                # etc.
[1] 2

Taking input as a vector instead gives a neater output format, although then the inputs are not technically lists.

JavaScript, 36 bytes

a=>[...a,0].map((x,y)=>a.slice(0,y))

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Mathematica, 22 21 bytes

^{-1 byte thanks to Misha Lavrov!}

{}~FoldList@Append~#&

Pure function. Takes a list as input and returns a list of lists as output. I believe this is the shortest possible solution.

Husk, 2 bytes

Θḣ

Gets all the ḣeads and then prepends Θ (in this case []):

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^{(needs type annotation for empty list: Try it online!)}

J, 5 bytes

a:,<\

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PowerShell, 65 bytes

param($a)'';$x=,0*($y=$a.count);0..--$y|%{$x[$_]=@($a[0..$_])};$x

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PowerShell helpfully unrolls lists-of-lists when the default Write-Output happens at program completion, so you get one item per line. Tack on a -join',' to see the list-of-lists better, by converting the inner lists into strings.

(Ab)uses the fact that attempting to output an empty array (e.g., @()) results in no output, so an empty array input just has '' as the output, since the $a[0..$_] will result in nothing. It will also throw out some spectacular error messages.

K (ngn/k), 8 bytes

,\(,!0),

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Common Lisp , 39 байт

(defun f(l)`(,@(if l(f(butlast l))),l))

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объяснение

(defun f(l)                           )  ; Define a function f
           `(                        )   ; With the list (essentially capable of interpolation), containing:
             ,@                          ;     The value of, flattened to one level
               (if l              )      ;         If l is not the empty list (which is the representation of nil, i.e. the only falsy value)
                    (f(butlast l))       ;         Recurse with all of l but the tail
                                   ,l    ;     The value of l

F #, 53 байта

На самом деле у меня есть два довольно похожих ответа, оба одинаковой длины. Они оба принимают общую последовательностьs as a parameter.

Первое решение:

let i s=Seq.init(Seq.length s+1)(fun n->Seq.take n s)

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Seq.takeзанимает первые nэлементы последовательности. Seq.initсоздает новую последовательность с количеством (в данном случае) длины последовательности sплюс 1, и для каждого элемента в последовательности принимает первые nэлементы в s.

Второе решение:

let i s=Seq.map(fun n->Seq.take n s){0..Seq.length s}

Similar to before, except it creates a sequence from 0 to the length of s. Then takes that number of elements from s.

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MATL, 15 12 bytes

3 bytes saved thanks to @Giuseppe

vin:"G@:)]Xh

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Due to the way that MATL displays the output, you can't explicitly see the empty array in the cell array. Here is a version that shows the output a little more explicitly.

Explanation

v       # Vertically concatenate the (empty) stack to create the array []
i       # Explicitly grab the input
n       # Compute the number of elements in the input (N)
:       # Create an array from [1, ..., N]
"       # Loop through this array
  G     # For each of these numbers, M
  @:    # Create an array from [1, ..., M]
  )     # Use this to index into the initial array
]       # End of the for loop
Xh      # Concatenate the entire stack into a cell array

SWI PROLOG 22 bytes

i(X,Y):-append(X,_,Y).

Charcoal, 6 bytes

Ｅθ…θκθ

Try it online! Link is to verbose version of code. Explanation:

 θ      Input array
Ｅ       Map over elements
   θ    Input array
  …     Moulded to length
    κ   Current loop index
        Implicitly print each array double-spaced
     θ  Input array
        Implicitly print

It's possible at a cost of 1 byte to ask Charcoal to print an n+1-element array which includes the input as its last element, but the output is the same, although the cursor position would be different if you then went on to print something else.

05AB1E, 3 bytes

η¯š

Explanation:

η    Prefixes
  š  Prepend
 ¯   Global array (empty by default)

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RAD, 7 bytes

(⊂⍬),,\

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This also works in Dyalog APL as a function.

How?

This works the same for both APL and RAD, given their close relation.

• (⊂⍬) the empty array
• , prepended to
• ,\ the prefixes (which exclude the empty array.)

Groovy, 37 bytes

{x->(0..x.size()).collect{x[0..<it]}}

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Japt, 5 bytes

Êò@¯Y

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brainfuck, 43 bytes

Take a list of non-null characters as input and returns all prefixes separated by newline. Requires double-infinite or wrapping tape.

,>-[+>,]<[-<]<<++++++++++[[<]>[.>]>[-<+>]<]

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C# (Visual C# Interactive Compiler), 39 bytes

x=>x.Select((_,i)=>x.Take(i)).Append(x)

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F# (Mono), 45 bytes

fun x->List.mapi(fun i y->List.take i x)x@[x]

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I am not totally sure if this is valid, but it seems like it follows the same "anonymous lambda" syntax that I've seem used in several other languages.

Java 8+, 86 77 bytes

^{-9 bytes thanks to Kevin Cruijssen (getting rid of the import)!}

x->java.util.stream.IntStream.range(0,x.size()+1).mapToObj(t->x.subList(0,t))

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Alternative, 65 bytes

The following will print the results to stdout (due to Olivier Grégoire):

x->{for(int i=0;i<=x.size();)System.out.print(x.subList(0,i++));}

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Brachylog, 9 bytes

a₀ᶠ~b.hĖ∧

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Explanation

a₀ᶠ           Find all prefixes of the input
   ~b         Add an element at the beginning of that list of prefixes
      hĖ      This element is the empty list
     .  ∧     (the list with the additional empty list is the output)

Ruby, 31 29 bytes

->a{[a*i=0]+a.map{a[0,i+=1]}}

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Explanation:

->a{             # take array input a
  [a*i=0]+       # set i to 0 and add whatever comes next to [[]] (a*0 == [])
  a.map{         # for every element in a (basically do a.length times)
    a[0,i+=1]  # increment i and return the first i-1 elements of a to map
  }
}