Code golf: combining multiple sorted lists into a

Question

Implement an algorithm to merge an arbitrary number of sorted lists into one sorted list. The aim is to create the smallest working programme, in whatever language you like.

For example:

input:  ((1, 4, 7), (2, 5, 8), (3, 6, 9))
output: (1, 2, 3, 4, 5, 6, 7, 8, 9)

input:  ((1, 10), (), (2, 5, 6, 7))
output: (1, 2, 5, 6, 7, 10)

Note: solutions which concatenate the input lists then use a language-provided sort function are not in-keeping with the spirit of golf, and will not be accepted:

sorted(sum(lists,[])) # cheating: out of bounds!

Apart from anything else, your algorithm should be (but doesn't have to be) a lot faster!

Clearly state the language, any foibles and the character count. Only include meaningful characters in the count, but feel free to add whitespace to the code for artistic / readability purposes.

To keep things tidy, suggest improvement in comments or by editing answers where appropriate, rather than creating a new answer for each "revision".

EDIT: if I was submitting this question again, I would expand on the "no language provided sort" rule to be "don't concatenate all the lists then sort the result". Existing entries which do concatenate-then-sort are actually very interesting and compact, so I won't retro-actively introduce a rule they break, but feel free to work to the more restrictive spec in new submissions.

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Inspired by Combining two sorted lists in Python

Answer 1

Though I have not had the patience to try this, a colleague of mine showed me a way that it may be possible to do this using 0 character key - Whie Space

Answer 2

Perl: 22 characters, including two significant whitespace characters.

sub a{sort map{@$_}@_}

Only builtins here. See? ;)

Call like so:

my @sorted = a([1, 2, 3], [5, 6, 89], [13, -1, 3]);
print "@sorted" # prints -1, 1, 1, 2, 3, 3, 5, 6, 89

Honestly, denying language features (note: not libraries...) seems kind-of counter the point. Shortest code to implement in a language should include buildins/language features. Of course, if you import a module, you should count that code against your solution.

Edit: removed unnecessary {}'s around the $_.

Answer 3

GNU system scripting (I guess that's cheating, but it is nice to know too).

sort -m file1 file2 file3 ...

Answer 4

BASH in about 250 essential chars

BASH is not really good at list manipulation, anyway this is does the job.

# This merges two lists together
m(){ 
    [[ -z $1 ]] && echo $2 && return; 
    [[ -z $2 ]] && echo $1 && return; 
    A=($1); B=($2); 
    if (( ${A[0]} > ${B[0]} ));then 
        echo -n ${B[0]}\ ;
        unset B[0];
    else 
        echo -n ${A[0]}\ ;
        unset A[0];
    fi;
    m "${A[*]}" "${B[*]}";
}
# This merges multiple lists
M(){
    A=$1;
    shift;
    for x in $@; do
        A=`m "$A" "$x"`
    done
    echo $A
}

$ M '1 4 7' '2 5 8' '3 6 9'
1 2 3 4 5 6 7 8 9

Answer 5

VB

The setup:

Sub Main()
    f(New Int32() {1, 4, 7}, _
      New Int32() {2, 5, 8}, _
      New Int32() {3, 6, 9})
End Sub

The output:

Sub f(ByVal ParamArray b As Int32()())
    Dim l = New List(Of Int32)
    For Each a In b
        l.AddRange(a)
    Next
    For Each a In l.OrderBy(Function(i) i)
        Console.WriteLine(a)
    Next
End Sub

Answer 6

Ruby: 100 characters (1 significant whitespace, 4 significant newlines)

def m(i)
  a=[]
  i.each{|s|s.each{|n|a.insert((a.index(a.select{|j|j>n}.last)||-1)+1,n)}}
  a.reverse
end

Human version:

def sorted_join(numbers)
  sorted_numbers=[]

  numbers.each do |sub_numbers|
    sub_numbers.each do |number|
      bigger_than_me = sorted_numbers.select { |i| i > number }
      if bigger_than_me.last
        pos = sorted_numbers.index(bigger_than_me.last) + 1
      else
        pos = 0
      end

      sorted_numbers.insert(pos, number)
    end
  end

  sorted_numbers.reverse
end

This can all just be replaced by numbers.flatten.sort

Benchmarks:

 a = [[1, 4, 7], [2, 4, 8], [3, 6, 9]]
 n = 50000
 Benchmark.bm do |b|
   b.report { n.times { m(a) } }
   b.report { n.times { a.flatten.sort } }
 end

Produces:

      user     system      total        real
 2.940000   0.380000   3.320000 (  7.573263)
 0.380000   0.000000   0.380000 (  0.892291)

So my algorithm performs horribly, yey!