挑战很简单，就是返回一个整数列表的列表，类似于Python range函数，只是每个连续的数字必须深于列表。

规则：

• 创建程序或非匿名函数
• 它应该返回或打印结果
• 结果应以列表（列表）或数组（数组）返回
• 如果参数为零，则返回一个空列表
• 这应该能够处理整数参数0 <= n <70。
  • （递归解决方案很快就爆炸了）
• 该函数只能使用一个参数调用。
• 其他行为是不确定的。
• 这是代码高尔夫，所以最短的代码获胜。

呼叫范例：

rangeList(6)
> [0, [1, [2, [3, [4, [5]]]]]]

测试用例：

0  => []
1  => [0]
2  => [0, [1]]
6  => [0, [1, [2, [3, [4, [5]]]]]]
26 => [0, [1, [2, [3, [4, [5, [6, [7, [8, [9, [10, [11, [12, [13, [14, [15, [16, [17, [18, [19, [20, [21, [22, [23, [24, [25]]]]]]]]]]]]]]]]]]]]]]]]]]
69 => [0, [1, [2, [3, [4, [5, [6, [7, [8, [9, [10, [11, [12, [13, [14, [15, [16, [17, [18, [19, [20, [21, [22, [23, [24, [25, [26, [27, [28, [29, [30, [31, [32, [33, [34, [35, [36, [37, [38, [39, [40, [41, [42, [43, [44, [45, [46, [47, [48, [49, [50, [51, [52, [53, [54, [55, [56, [57, [58, [59, [60, [61, [62, [63, [64, [65, [66, [67, [68]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]

编辑： isaacg的答案是迄今为止最短的答案。如果有人用挑战发布时所用的语言找到了一个简短的答案，我将更新接受的答案。感谢参与！

Pyth，13个字节

?hu]+HG_UQYQY

在这里尝试。

                 Implicit:
                 Q = eval(input())
                 Y = []
?           QY   If Q = 0, print Y
 h               else, print the first element of
  u     _UQY     the reduce, where Y is the initial value, over the list
                 reversed(range(Q))
   ]+HG          The reduce function: The list containing H prepended onto G.
                 The new number is inserted into the accumulated list,
                 then the resultant list is wrapped in another list.

杀伤人员地雷（13 18）

假设⎕IO=0：

f←{×⍵:⊃,∘⊂∘,/⍳⍵⋄⍬}

说明：

• ×⍵:如果⍵是肯定的，
  • ,∘⊂∘,：将左侧操作数加入到右侧操作数的包围中（例如x ,∘⊂∘, y = [x, [y]]）
  • /： 降低
  • ⍳⍵： 数字 0..⍵-1
  • ⊃：披露结果
• ⋄： 除此以外
  • ⍬：返回空列表
  • （这是必需的，因为/在上失败⍬，并⍳0给出了空列表。）

附录：

该函数返回一个嵌套数组。但是，从APL的默认输出中很难分辨出这一点。它通过空格分隔数组项，因此您只能通过双空格来区分嵌套。这是一个函数，该函数将使用嵌套数组并返回一个字符串，以Python样式（即[a,[b,[c,...]]]）格式化嵌套数组。

arrfmt←{0=≡⍵:⍕⍵ ⋄ '[',(1↓∊',',¨∇¨⍵),']'}

Haskell，67个字节

data L=E|I Int|L[L] 
1#m=L[I$m-1]
n#m=L[I$m-n,(n-1)#m]
p 0=E
p n=n#n

在Haskell中，列表的所有元素必须具有相同的类型，因此我不能将整数与整数列表混合使用，而必须定义自定义列表类型L。helper函数以#递归方式构造所需列表。main函数p检查空列表，#否则调用。

由于默认情况下无法打印新的数据类型（规则仅允许返回列表），因此我添加了一些代码进行演示：

data L=E|I Int|L[L] deriving Show

现在：

-- mapM_ (print . p) [0..5]
E
L [I 0]
L [I 0,L [I 1]]
L [I 0,L [I 1,L [I 2]]]
L [I 0,L [I 1,L [I 2,L [I 3]]]]
L [I 0,L [I 1,L [I 2,L [I 3,L [I 4]]]]]

Python，48个字节

f=lambda n,i=0:i<n and[i]+[f(n,i+1)]*(i<n-1)or[]

使用列表乘法处理特殊情况。

Mathematica，33岁

f@0={};f@1={0};f@n_:={0,f[n-1]+1}

CJam，16个字节

Lri){[}%]~;']*~p

这是一个完整程序。它通过STDIN接收输入，并将最终数组打印到STDOUT上。

与其他CJam条目一样，0输入将被打印，""因为这是CJam中空数组的表示。

运作方式：

L                   "Put an empty array on stack. This will be used for the 0 input";
 ri)                "Read the input, convert it to integer and increment it";
    {[}%            "Map over the array [0 ... input number] starting another array";
                    "after each element";
        ]~;         "Now on stack, we have input number, an empty array and the final";
                    "opening bracket. Close that array, unwrap it and pop the empty array";
           ']*~     "Put a string containing input number of ] characters and eval it";
                    "This closes all the opened arrays in the map earlier";
               p    "Print the string representation of the array";
                    "If the input was 0, the map runs 1 time and the ; pops that 1 array";
                    "Thus leaving only the initial empty array on stack";

在这里在线尝试

JavaScript（ES6）40

递归解决方案，非常健壮，没有任何打击。 6500附近更新失败，“递归过多”

F=n=>n--?(R=m=>m<n?[m,R(++m)]:[m])(0):[]

迭代解决方案（45）除内存使用量外没有其他限制

F=n=>{for(s=n?[--n]:[];n;)s=[--n,s];return s}

尝试F（1000）：FireBug控制台不会向您显示超过190个嵌套数组，但是它们在那里

Java中，88个 107 105 104 102字节

import java.util.*;int o;List f(final int n){return new Stack(){{add(n<1?"":o++);if(o<n)add(f(n));}};}

与其他语言相比，它相当长，尽管Java不能做得更好。仅需检查即可确定是否继续递归。

Python 2，56个字节

我怀疑这可能会打更多。

f=lambda n,i=0:[i,f(n,i+1)]if i<n-1 else[i]if n>0 else[]

测试：

# for n in (0,1,2,6,26,69): print n, '=>', f(n)
0 => []
1 => [0]
2 => [0, [1]]
6 => [0, [1, [2, [3, [4, [5]]]]]]
26 => [0, [1, [2, [3, [4, [5, [6, [7, [8, [9, [10, [11, [12, [13, [14, [15, [16, [17, [18, [19, [20, [21, [22, [23, [24, [25]]]]]]]]]]]]]]]]]]]]]]]]]]
69 => [0, [1, [2, [3, [4, [5, [6, [7, [8, [9, [10, [11, [12, [13, [14, [15, [16, [17, [18, [19, [20, [21, [22, [23, [24, [25, [26, [27, [28, [29, [30, [31, [32, [33, [34, [35, [36, [37, [38, [39, [40, [41, [42, [43, [44, [45, [46, [47, [48, [49, [50, [51, [52, [53, [54, [55, [56, [57, [58, [59, [60, [61, [62, [63, [64, [65, [66, [67, [68]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]

CJam，17个字节

我知道Optimizer找到了16个，但是这是我能做的最好的：

{:I{[}%;{]}I1e>*}

这是一个块，与CJam中的函数最接近，它在堆栈上采用整数，并保留所需的嵌套数组。

使用此程序对其进行测试，它将输入放入堆栈中，然后调用该函数并检查堆栈。请注意，对于0，堆栈输出将包含""-这是 CJam对空数组的本机表示。

红宝石46

f=->n{r=a=[]
(0...n).map{|i|a<<a=[i]}
r[0]||r}

在线测试：http：//ideone.com/uYRVTa

C＃-100

简单递归。检查零的特殊情况，并用一个变量打勾，用另一个变量打勾

object[]A(int y,int x=0){return y==0?new object[0]:y==1?new object[]{x}:new object[]{x,A(--y,++x)};}

C ++ 87

（Visual C ++ 2012）

int*A(int y,int x=0){int*b=new int{x};return!y?new int:--y?(b[1]=(int)A(y,++x))?b:0:b;}

这个很棒，我的意思是拜占庭，但是它与c＃的基本思想相同。

它是C样式的数组实现，因此它不给您一个数组，而是提供一个int指针，在其中我同时存储了int和其他指针。像这样：[0,*] *->[1,#] #-> [2,&] &-> etc，其中符号是指针的int值的伪代码，而->是指针在内存中的指针。

我已经设计（咳嗽）了一个非常好用的c样式锯齿状数组的实现，但是我认为它足够合理，可以放在问题的规则之内。

这里有很多滥用三元运算符，也有很多滥用从int到bool的隐式强制转换。

示例：如果让int *bar = (int*)A(3);，我们可以看到：

bar
0x003bded8 {0}
((int*)bar[1])[0]
1
((int*)(((int*)bar[1])[1]))[0]
2

这是[0，[1，[2]]]的指针对话。

好的。实际上并不一定要很糟糕。这是一些运行此c ++代码的测试代码：

int* GetNext(int* p){
  return (int*)p[1];
}

int main()
{
    auto x = 10;
    auto bar = A(x);

    for (int i = 1; i < x; i++){
        bar = GetNext(bar);
        std::cout << bar[0] << std::endl;
    }

}

Pyth，15个字节

?u[HG)_UtQ]tQQY

真正的意思是，在Python中：

Q = eval(input())
if Q:
    print reduce(lambda G,H:[H,G], reverse(range(Q-1)), [Q-1])
else:
    print []

Haskell中，65个59 45 41字节

这些嵌套列表与root Tree的数据结构相同，不同之处在于它们也可以为空。因此，我们可以使用它们的列表-也称为a Forest来代表它们。

(0!)
data T=N[T]Int
m!n=[N((m+1)!n)m|m<n]

在线尝试！

说明

首先我们需要实现Tree数据类型：

data Tree = Node [Tree] Int

从那里开始，仅使用两个参数m（递增计数）进行递归并n跟踪何时终止：

m ! n= [ Node ((m+1)!n) m| m<n ]

备用，61字节

import Data.Tree
f n=unfoldForest(\b->(b,[b+1|b<n-1]))[0|n>0]

在线尝试！

说明

该函数unfoldForest采用一个初始值列表和一个函数x -> (y,[x])。对于每个初始值，x它使用函数展开一棵树，生成一个元组(y,xs)，该元组y将成为根，xs并用于重复该过程：

unfoldForest (\b -> (b, [b+1 | b < 2]) [0]
  ≡ Node 0 [unfoldForest (\b -> (b, [b+1 | b < 2) [1]]
  ≡ Node 0 [Node 1 [unfoldForest (\b -> (b, [b+1 | b < 2) []]]
  ≡ Node 0 [Node 1 []]

Perl-44

sub t{$r=[($t)=@_];$r=[$t,$r]while--$t>0;$r}

将根据要求添加说明。您可以在这里尝试。

JavaScript，93个字节

这不是理想的选择，但我不妨尝试一下。稍后，我将尝试进一步打高尔夫球，尽管到目前为止，我还没有找到明显的方法。

function f(n){s='[';i=0;while(i<n-1)s+=i+++',[';s+=i||'';do{s+=']'}while(i--);return eval(s)}

Python，75个字节

这只是为了展示。这是我在创建/设计挑战时编写的程序。

f=lambda x,y=[]:y if x<1 else f(x-1,[x-2]+[y or[x-1]])if x>1 else y or[x-1]

Python，44岁

f=lambda n,i=0:i<n-1and[i,f(n,i+1)]or[i][:n]

递归创建树。将[:n]在年底是特殊情况下的n==0进给空单。

乔 8字节

注意：这是非竞争性答案。在此问题之后，发布了Joe的第一个版本。

F:/+,M]R

我们有什么在这里？F:定义了一个函数F是的链/+,，M]和R。调用时Fn，首先Rn被求值，返回范围从0到n（不包括）。M]将每个元素包装到一个列表中。然后将该列表应用于/+,。x +, y 返回x + [y]。/是正确的选择。因此，/+,a b c d...return [a, [b, [c, [d...]]]。

调用示例（代码缩进3，输出缩进0）：

   F:/+,M]R
   F10
[0, [1, [2, [3, [4, [5, [6, [7, [8, [9]]]]]]]]]]
   F2
[0, [1]]
   F1
[0]
   F0
[]
   F_5
[0, [-1, [-2, [-3, [-4]]]]]

Ruby-递归版本-52

r=->(n,v=nil){(n-=1;n<0 ?v:r[n,(v ?[n,v]:[n])])||[]}

非递归版本：66 62 57

r=->i{(i-1).downto(0).inject(nil){|a,n|a ?[n,a]:[n]}||[]}

样本输出（两个版本相同）

p r[0]  # => []
p r[1]  # => [0]
p r[2]  # => [0, [1]]
p r[6]  # => [0, [1, [2, [3, [4, [5]]]]]]
p r[26] # => [0, [1, [2, [3, [4, [5, [6, [7, [8, [9, [10, [11, [12, [13, [14, [15, [16, [17, [18, [19, [20, [21, [22, [23, [24, [25]]]]]]]]]]]]]]]]]]]]]]]]]]
p r[69] # => [0, [1, [2, [3, [4, [5, [6, [7, [8, [9, [10, [11, [12, [13, [14, [15, [16, [17, [18, [19, [20, [21, [22, [23, [24, [25, [26, [27, [28, [29, [30, [31, [32, [33, [34, [35, [36, [37, [38, [39, [40, [41, [42, [43, [44, [45, [46, [47, [48, [49, [50, [51, [52, [53, [54, [55, [56, [57, [58, [59, [60, [61, [62, [63, [64, [65, [66, [67, [68]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]

非递归版本可以处理任意大的输入。

p r[1000] # => [0, [1, [2, [3, [4, [5, [6, [7, [8, [9, [10, [11, [12, [13, [14, [15, [16, [17, [18, [19, [20, [21, [22, [23, [24, [25, [26, [27, [28, [29, [30, [31, [32, [33, [34, [35, [36, [37, [38, [39, [40, [41, [42, [43, [44, [45, [46, [47, [48, [49, [50, [51, [52, [53, [54, [55, [56, [57, [58, [59, [60, [61, [62, [63, [64, [65, [66, [67, [68, [69, [70, [71, [72, [73, [74, [75, [76, [77, [78, [79, [80, [81, [82, [83, [84, [85, [86, [87, [88, [89, [90, [91, [92, [93, [94, [95, [96, [97, [98, [99, [100, [101, [102, [103, [104, [105, [106, [107, [108, [109, [110, [111, [112, [113, [114, [115, [116, [117, [118, [119, [120, [121, [122, [123, [124, [125, [126, [127, [128, [129, [130, [131, [132, [133, [134, [135, [136, [137, [138, [139, [140, [141, [142, [143, [144, [145, [146, [147, [148, [149, [150, [151, [152, [153, [154, [155, [156, [157, [158, [159, [160, [161, [162, [163, [164, [165, [166, [167, [168, [169, [170, [171, [172, [173, [174, [175, [176, [177, [178, [179, [180, [181, [182, [183, [184, [185, [186, [187, [188, [189, [190, [191, [192, [193, [194, [195, [196, [197, [198, [199, [200, [201, [202, [203, [204, [205, [206, [207, [208, [209, [210, [211, [212, [213, [214, [215, [216, [217, [218, [219, [220, [221, [222, [223, [224, [225, [226, [227, [228, [229, [230, [231, [232, [233, [234, [235, [236, [237, [238, [239, [240, [241, [242, [243, [244, [245, [246, [247, [248, [249, [250, [251, [252, [253, [254, [255, [256, [257, [258, [259, [260, [261, [262, [263, [264, [265, [266, [267, [268, [269, [270, [271, [272, [273, [274, [275, [276, [277, [278, [279, [280, [281, [282, [283, [284, [285, [286, [287, [288, [289, [290, [291, [292, [293, [294, [295, [296, [297, [298, [299, [300, [301, [302, [303, [304, [305, [306, [307, [308, [309, [310, [311, [312, [313, [314, [315, [316, [317, [318, [319, [320, [321, [322, [323, [324, [325, [326, [327, [328, [329, [330, [331, [332, [333, [334, [335, [336, [337, [338, [339, [340, [341, [342, [343, [344, [345, [346, [347, [348, [349, [350, [351, [352, [353, [354, [355, [356, [357, [358, [359, [360, [361, [362, [363, [364, [365, [366, [367, [368, [369, [370, [371, [372, [373, [374, [375, [376, [377, [378, [379, [380, [381, [382, [383, [384, [385, [386, [387, [388, [389, [390, [391, [392, [393, [394, [395, [396, [397, [398, [399, [400, [401, [402, [403, [404, [405, [406, [407, [408, [409, [410, [411, [412, [413, [414, [415, [416, [417, [418, [419, [420, [421, [422, [423, [424, [425, [426, [427, [428, [429, [430, [431, [432, [433, [434, [435, [436, [437, [438, [439, [440, [441, [442, [443, [444, [445, [446, [447, [448, [449, [450, [451, [452, [453, [454, [455, [456, [457, [458, [459, [460, [461, [462, [463, [464, [465, [466, [467, [468, [469, [470, [471, [472, [473, [474, [475, [476, [477, [478, [479, [480, [481, [482, [483, [484, [485, [486, [487, [488, [489, [490, [491, [492, [493, [494, [495, [496, [497, [498, [499, [500, [501, [502, [503, [504, [505, [506, [507, [508, [509, [510, [511, [512, [513, [514, [515, [516, [517, [518, [519, [520, [521, [522, [523, [524, [525, [526, [527, [528, [529, [530, [531, [532, [533, [534, [535, [536, [537, [538, [539, [540, [541, [542, [543, [544, [545, [546, [547, [548, [549, [550, [551, [552, [553, [554, [555, [556, [557, [558, [559, [560, [561, [562, [563, [564, [565, [566, [567, [568, [569, [570, [571, [572, [573, [574, [575, [576, [577, [578, [579, [580, [581, [582, [583, [584, [585, [586, [587, [588, [589, [590, [591, [592, [593, [594, [595, [596, [597, [598, [599, [600, [601, [602, [603, [604, [605, [606, [607, [608, [609, [610, [611, [612, [613, [614, [615, [616, [617, [618, [619, [620, [621, [622, [623, [624, [625, [626, [627, [628, [629, [630, [631, [632, [633, [634, [635, [636, [637, [638, [639, [640, [641, [642, [643, [644, [645, [646, [647, [648, [649, [650, [651, [652, [653, [654, [655, [656, [657, [658, [659, [660, [661, [662, [663, [664, [665, [666, [667, [668, [669, [670, [671, [672, [673, [674, [675, [676, [677, [678, [679, 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[999]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]

两个版本也都接受负数

p r[-5] # => []

PHP 5.4（67字节）：

我知道我知道。

这绝不是最短的答案。

但这有效！

这里是：

function F($n){for($c=$n?[--$n]:[];~$n&&$n--;$c=[$n,$c]);return$c;}

您可以在这里进行测试：https：//ideone.com/42L35E（忽略错误）

Javascript（57位元组）：

这是相同的 确切 代码除了JavaScript是挑剔的回报，我已经减少了变量名：

function F(n){for(c=n?[--n]:[];~n&&n--;c=[n,c]);return c}

看到？相同的代码！

ES6（49个字节）：

基本相同的完全相同的代码，但针对ES6：

F=n=>{for(c=n?[--n]:[];~n&&n--;c=[n,c]);return c}

Javascript（114位元组）：

其他人都在递归，所以我想尝试一个迭代解决方案。不过，我有太多特殊情况。

我拿着一个主列表，然后循环并用新数字附加新列表。

function q(n){a=[];if(n==1)a=[0];else if(n!=0){a=[0,b=[]];for(i=1;i<n;){c=[];b.push(c);b.push(i++);b=c}}return a}

通用Lisp（95字节）：

(defun f(n &optional(i 0))(cond((< i(1- n))(cons i(list(f n(1+ i)))))((> n 0)(list i))(t nil)))

JavaScript，35 37字节

递归解

f=(n,x=[])=>n?f(--n,x[0]?[n,x]:[n]):x

在线尝试！

05AB1E，11 个字节

_i¯ëFNI<α)R

在线尝试或验证所有测试用例。

11字节替代：

_i¯ëݨRvy)R

在线尝试或验证所有测试用例。

说明：

05AB1E没有向下的循环，因此要在范围内循环，(input, 0]我必须：

• 首先创建该范围（ݨR；创建范围[0, input]，删除最后一项，反向），然后在其上循环（vy）；
• 或在范围内循环[0, input)（F），并求出循环索引与输入1（NI<α）之间的绝对差。

_i          # If the (implicit) input is 0:
  ¯         #  Push the global array (empty by default)
 ë          # Else:
  F         #  Loop `N` in the range [0, (implicit) input):
   N        #   Push index `N`
    I<      #   Push input-1
      α     #   Take the absolute difference between the two
       )    #   Wrap everything on the stack into a list
        R   #   Reverse the list
            # (output the result implicitly after the if/loop)