您的任务很简单：

给定整数集列表，找到集合并集。换句话说，找到包含原始集列表中的所有元素（但没有其他元素）的整数集的最短列表。例如：

[1,5] and [3,9] becomes [1,9] as it contains all of the elements in both [1,5] and [3,9]
[1,3] and [5,9] stays as [1,3] and [5,9], because you don't want to include 4

集使用范围表示法表示：[1,4]表示整数1,2,3,4。集也可以是无界的：[3,]表示所有整数>= 3，并且[,-1]表示所有整数<= -1。确保范围的第一个元素不大于第二个。

您可以选择采用字符串符号表示的集合，也可以使用2元素元组，并使用常量非整数作为“无限”值。您可以使用两个不同的常量来表示无限的上限和无限的下限。例如，在Javascript中，只要您在所有测试用例中都使用一致，就可以使用它[3,{}]来标记所有整数。>= 3{}

测试用例：

[1,3]                     => [1,3]
[1,]                      => [1,]
[,9]                      => [,9]
[,]                       => [,]
[1,3],[4,9]               => [1,9]
[1,5],[8,9]               => [1,5],[8,9]
[1,5],[1,5]               => [1,5]
[1,5],[3,7]               => [1,7]
[-10,7],[1,5]             => [-10,7]
[1,1],[2,2],[3,3]         => [1,3]
[3,7],[1,5]               => [1,7]
[1,4],[8,]                => [1,4],[8,]
[1,4],[-1,]               => [-1,]
[1,4],[,5]                => [,5]
[1,4],[,-10]              => [1,4],[,-10]
[1,4],[,]                 => [,]
[1,4],[3,7],[8,9],[11,20] => [1,9],[11,20]

这是代码高尔夫，因此请尽可能短地回答！

R + intervals，90 87 81字节

function(...)t(reduce(Intervals(rbind(...),type="Z")))+c(1,-1)
library(intervals)

在线尝试！

输入是间隔列表。-Inf并且Inf是R内置的负/正无穷大。输出是间隔列的矩阵。

通常不喜欢使用非标准库，但这很有趣。TIO尚未intervals安装。您可以在自己的安装中或在https://rdrr.io/snippets/上尝试

该intervals程序包支持实数和整数（type = "Z"）间隔，并且该reduce函数是挑战所需内容的内置函数，但是输出似乎默认为打开间隔，因此需要此函数才能获得所需的结果。close_intervals +c(1,-1)

旧版本在列表列表中有示例，可能会很方便，因此我在此处留下了链接。

JavaScript（ES6），103个字节

感谢@Shaggy 保存了1个字节感谢@KevinCruijssen
保存了1个字节

预计+/-Infinity无限价值。

a=>(a.sort(([p],[P])=>p-P).map(m=M=([p,q])=>p<M+2?M=q>M?q:M:(b.push([m,M]),m=p,M=q),b=[]),b[0]=[m,M],b)

在线尝试！

怎么样？

我们首先按照区间的下限（从最低到最高）对区间进行排序。上限被忽略。

然后，我们在排序的时间间隔$[p_n,q_n]$进行迭代，同时跟踪当前的上下限$m$和$M$，分别将其初始化为$p_1$和$q_1$。

对于每个间隔$[p_n,q_n]$：

• 如果$p_n\le M+1$：这个间隔可以与前面的合并。但是我们可能有一个新的上限，因此我们将$M$更新为$\max(M,q_n)$。
• $[m,M]$$m$$M$$p_n$$q_n$

$[m,M]$

已评论

a => (                  // a[] = input array
  a.sort(([p], [P]) =>  // sort the intervals by their lower bound; we do not care about
    p - P)              // the upper bounds for now
  .map(m = M =          // initialize m and M to non-numeric values
    ([p, q]) =>         // for each interval [p, q] in a[]:
    p < M + 2 ?         //   if M is a number and p is less than or equal to M + 1:
      M = q > M ? q : M //     update the maximum M to max(M, q)
    : (                 //   else (we've found a gap, or this is the 1st iteration):
      b.push([m, M]),   //     push the interval [m, M] in b[]
      m = p,            //     update the minimum m to p
      M = q             //     update the maximum M to q
    ),                  //
    b = []              //   start with b[] = empty array
  ),                    // end of map()
  b[0] = [m, M], b      // overwrite the 1st entry of b[] with the last interval [m, M]
)                       // and return the final result

Python 2中，118 113 112 111 106 105个 104 101字节

x=input()
x.sort();a=[];b,c=x[0]
for l,r in x:
 if l>c+1:a+=(b,c),;b,c=l,r
 c=max(c,r)
print[(b,c)]+a

多亏了Xcoder先生，节省了一个字节，感谢Jonathan Frech节省了一个字节，感谢Dead Possum节省了三个字节。
在线尝试！

红宝石，89 76字节

->a{[*a.sort.reduce{|s,y|s+=y;y[0]-s[-3]<2&&s[-3,3]=s.max;s}.each_slice(2)]}

在线尝试！

对数组进行排序，然后通过将所有范围附加到第一个范围进行展平：如果范围与前一个范围重叠，则从后三个范围中丢弃2个元素（仅保留最大值）。

最后展开所有内容。

帕斯卡（FPC） ，367个 362 357字节

uses math;type r=record a,b:real end;procedure d(a:array of r);var i,j:word;t:r;begin for i:=0to length(a)-1do for j:=0to i do if a[i].a<a[j].a then begin t:=a[i];a[i]:=a[j];a[j]:=t;end;j:=0;for i:=1to length(a)-1do if a[j].b>=a[i].a-1then begin a[j].a:=min(a[i].a,a[j].a);a[j].b:=max(a[i].b,a[j].b)end else j:=j+1;for i:=0to j do writeln(a[i].a,a[i].b)end;

在线尝试！

该过程采用由2个范围边界组成的动态记录数组，对数组进行适当修改，然后将其写入标准输出，每行一个范围。（很抱歉，这个扭曲的句子。）用于无界1/0向上和-1/0无界向下。

可读版本

仅返回具有正确数目的元素的数组会很好，但是传递给函数/过程的动态数组不再是动态数组...首先，我找到了这个，然后有一个很棒的，令人费解的解释。

这是我可以找到的用于缩短代码的最佳数据结构。如果您有更好的选择，请随时提出建议。

Wolfram语言（Mathematica），57个字节

List@@(#-{0,1}&/@IntervalUnion@@(Interval[#+{0,1}]&/@#))&

在线尝试！

将输入作为{a,b}代表时间间隔的列表的列表[a,b]，其中abe可以-Infinity和bbe可以Infinity。

使用内置的IntervalUnion，但我们当然必须先将间隔按摩成形状。为了假装间隔是整数，我们将1加到上限（确保[1,3]and的并集[4,9]是[1,9]）。最后，我们撤消该操作，然后将结果返回到列表列表。

还有一种完全不同的方法，它的时钟为73个字节：

NumericalSort@#//.{x___,{a_,b_},{c_,d_},y___}/;b+1>=c:>{x,{a,b~Max~d},y}&

在这里，在对间隔进行排序之后，只要将两个连续的间隔合并为一个间隔，便用它们的并集替换它们，然后重复进行直到没有剩下要进行的此类操作为止。

05AB1E（旧版），88 79 78 字节

g≠i˜AKïDW<UZ>VIøεAXY‚Nè:}ïø{©˜¦2ôíÆ1›.œʒíεćsO_*}P}н€g®£εø©θàDYQiA}V®нßDXQiA}Y‚

输入无穷大作为小写字母（'abcdefghijklmnopqrstuvwxyz'）。

在线尝试或验证所有测试用例。

重要说明：如果有一个实际的Infinity和-Infinity，则应该是43 42个字节。所以，略超过50％， 30％左右是作为变通的缺乏Infinity..

{©Dg≠i˜¦2ôíÆ1›.œʒíεćsO_*}P}н€g®£εø©θàV®нßY‚

在线试用（与Infinity替换为9999999999与-Infinity替换成-9999999999）。

绝对可以打高尔夫球。最后，结果发现非常非常丑陋的解决方法。但是现在我很高兴它正在工作。

说明：

Dg≠i          # If the length of the implicit input is NOT 1:
              #   i.e. [[1,3]] → length 1 → 0 (falsey)
              #   i.e. [[1,4],["a..z",-5],[3,7],[38,40],[8,9],[11,20],[25,"a..z"],[15,23]]
              #    → length 8 → 1 (truthy)
    ˜         #  Take the input implicit again, and flatten it
              #   i.e. [[1,4],["a..z",-5],[3,7],[38,40],[8,9],[11,20],[25,"a..z"],[15,23]]
              #    → [1,4,"a..z",-5,3,7,38,40,8,9,11,20,25,"a..z",15,23]
     AK       #  Remove the alphabet
              #   i.e. [1,4,"a..z",-5,3,7,38,40,8,9,11,20,25,"a..z",15,23]
              #    → ['1','4','-5','3','7','38','40','8','9','11','20','25','15','23']
       ï      #  Cast everything to an integer, because `K` turns them into strings..
              #   i.e. ['1','4','-5','3','7','38','40','8','9','11','20','25','15','23']
              #    → [1,4,-5,3,7,38,40,8,9,11,20,25,15,23]
        D     #  Duplicate it
         W<   #  Determine the min - 1
              #   i.e. [1,4,-5,3,7,38,40,8,9,11,20,25,15,23] → -5
           U  #  Pop and store it in variable `X`
         Z>   #  Determine the max + 1
              #   i.e. [1,4,-5,3,7,38,40,8,9,11,20,25,15,23] → 40
           V  #  Pop and store it in variable `Y`
Iø            #  Take the input again, and transpose/zip it (swapping rows and columns)
              #   i.e. [[1,4],["a..z",-5],[3,7],[38,40],[8,9],[11,20],[25,"a..z"],[15,23]]
              #    → [[1,'a..z',3,38,8,11,25,15],[4,-5,7,40,9,20,'a..z',23]]
  ε       }   #  Map both to:
   A          #   Push the lowercase alphabet
    XY‚       #   Push variables `X` and `Y`, and pair them into a list
       Nè     #   Index into this list, based on the index of the mapping
         :    #   Replace every alphabet with this min-1 or max+1
              #   i.e. [[1,'a..z',3,38,8,11,25,15],[4,-5,7,40,9,20,'a..z',23]]
              #    → [['1','-6','3','38','8','11','25','15'],['4','-5','7','40','9','20','41','23']]
ï             #  Cast everything to integers again, because `:` turns them into strings..
              #   i.e. [['1','-6','3','38','8','11','25','15'],['4','-5','7','40','9','20','41','23']]
              #    → [[1,-6,3,38,8,11,25,15],[4,-5,7,40,9,20,41,23]]
 ø            #  Now zip/transpose back again
              #   i.e. [[1,-6,3,38,8,11,25,15],[4,-5,7,40,9,20,41,23]]
              #    → [[1,4],[-6,-5],[3,7],[38,40],[8,9],[11,20],[25,41],[15,23]]
  {           #  Sort the pairs based on their lower range (the first number)
              #   i.e. [[1,4],[-6,-5],[3,7],[38,40],[8,9],[11,20],[25,41],[15,23]]
              #    → [[-6,-5],[1,4],[3,7],[8,9],[11,20],[15,23],[25,41],[38,40]]
   ©          #  Store it in the register (without popping)
˜             #  Flatten the list
              #   i.e. [[-6,-5],[1,4],[3,7],[8,9],[11,20],[15,23],[25,41],[38,40]]
              #    → [-6,-5,1,4,3,7,8,9,11,20,15,23,25,41,38,40]
 ¦            #  And remove the first item
              #   i.e. [-6,-5,1,4,3,7,8,9,11,20,15,23,25,41,38,40]
              #    → [-5,1,4,3,7,8,9,11,20,15,23,25,41,38,40]
  2ô          #  Then pair every two elements together
              #   i.e. [-5,1,4,3,7,8,9,11,20,15,23,25,41,38,40]
              #    → [[-5,1],[4,3],[7,8],[9,11],[20,15],[23,25],[41,38],[40]]
    í         #  Reverse each pair
              #   i.e. [[-5,1],[4,3],[7,8],[9,11],[20,15],[23,25],[41,38],[40]]
              #    → [[1,-5],[3,4],[8,7],[11,9],[15,20],[25,23],[38,41],[40]]
     Æ        #  Take the difference of each pair (by subtracting)
              #   i.e. [[1,-5],[3,4],[8,7],[11,9],[15,20],[25,23],[38,41],[40]]
              #    → [6,-1,1,2,-5,2,-3,40]
      1›      #  Determine for each if they're larger than 1
              #   i.e. [6,-1,1,2,-5,2,-3,40] → [1,0,0,1,0,1,0,1]
.œ            #  Create every possible partition of these values
              #   i.e. [1,0,0,1,0,1,0,1] → [[[1],[0],[0],[1],[0],[1],[0],[1]],
              #                             [[1],[0],[0],[1],[0],[1],[0,1]],
              #                             ...,
              #                             [[1,0,0,1,0,1,0],[1]],
              #                             [[1,0,0,1,0,1,0,1]]]
  ʒ         } #  Filter the partitions by:
   í          #   Reverse each inner partition
              #    i.e. [[1],[0,0,1],[0,1],[0,1]] → [[1],[1,0,0],[1,0],[1,0]]
    ε     }   #   Map each partition to:
     ć        #    Head extracted
              #     i.e. [1,0,0] → [0,0] and 1
              #     i.e. [1] → [] and 1
              #     i.e. [1,0,1] → [1,0] and 1
      s       #    Swap so the rest of the list is at the top of the stack again
       O      #    Take its sum
              #     i.e. [0,0] → 0
              #     i.e. [] → 0
              #     i.e. [1,0] → 1
        _     #    And check if it's exactly 0
              #     i.e. 0 → 1 (truthy)
              #     i.e. 1 → 0 (falsey)
         *    #    And multiply it with the extracted head
              #    (is only 1 when the partition has a single trailing 1 and everything else a 0)
              #     i.e. 1 and 1 → 1 (truthy)
              #     i.e. 1 and 0 → 0 (falsey)
           P  #   And check if all mapped partitions are 1
н             #  Take the head (there should only be one valid partition left)
              #   i.e. [[[1],[0,0,1],[0,1],[0,1]]] → [[1],[0,0,1],[0,1],[0,1]]
 €g           #  Take the length of each inner list
              #   i.e. [[1],[0,0,1],[0,1],[0,1]] → [1,3,2,2]
   ®          #  Push the sorted pairs we've saved in the register earlier
    £         #  Split the pairs into sizes equal to the partition-lengths
              #   i.e. [1,3,2,2] and [[-6,-5],[1,4],[3,7],[8,9],[11,20],[15,23],[25,41],[38,40]]
              #    → [[[-6,-5]],[[1,4],[3,7],[8,9]],[[11,20],[15,23]],[[25,41],[38,40]]]
ε             #  Map each list of pairs to:
 ø            #   Zip/transpose (swapping rows and columns)
              #    i.e. [[1,4],[3,7],[8,9]] → [[1,3,8],[4,7,9]]
              #    i.e. [[25,41],[38,40]] → [[25,38],[41,40]]
  ©           #   Store it in the register
   θ          #   Take the last list (the ending ranges)
              #    i.e. [[25,38],[41,40]] → [41,40]
    à         #   And determine the max
              #    i.e. [41,40] → 41
     DYQi }   #   If this max is equal to variable `Y`
              #     i.e. 41 (`Y` = 41) → 1 (truthy)
         A    #    Replace it back to the lowercase alphabet
           V  #   Store this max in variable `Y`
  ®           #   Take the zipped list from the register again
   н          #   This time take the first list (the starting ranges)
              #    i.e. [[25,38],[41,40]] → [25,38]
    ß         #   And determine the min
              #    i.e. [25,38] → 25
     DXQi }   #   If this min is equal to variable `X`
              #     i.e. 25 (`X` = -6) → 0 (falsey)
         A    #    Replace it back to the lowercase alphabet
           Y‚ #   And pair it up with variable `Y` (the max) to complete the mapping
              #    i.e. 25 and 'a..z' → [25,'a..z']
              #  Implicitly close the mapping (and output the result)
              #   i.e. [[[-6,-5]],[[1,4],[3,7],[8,9]],[[11,20],[15,23]],[[25,41],[38,40]]]
              #    → [['a..z',-5],[1,9],[11,23],[25,'a..z']]
              # Implicit else (only one pair in the input):
              #  Output the (implicit) input as is
              #   i.e. [[1,3]]

C（clang），346342字节

编译器标志-DP=printf("(%d,%d)\n"，-DB=a[i+1]和-DA=a[i]

typedef struct{int a,b;}t;s(t**x,t**y){if((*x)->a>(*y)->a)return 1;else if((*x)->a<(*y)->a)return -1;}i;f(t**a){for(i=0;A;)i++;qsort(a,i,sizeof(t*),s);for(i=0;B;i++){if(B->a<=A->b+1){A->b=B->b;if(B->a<A->a)A->a=B->a;else B->a=A->a;}}for(i=0;A;i++){if(!B)break;if(A->a!=B->a)P,A->a,A->b);}P,A->a,A->b);}

在线尝试！

Stax，46 39 字节

ÿδ7│ⁿ╚╪║»ÿ1Wç♥├óπ◙+╣╝[á╬p£ß₧ΓÑ°♥ºië«4T╗

运行并调试

该程序以原始指定的字符串表示形式输入。