Code golf: combining multiple sorted lists into a single sorted list

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.

Answer 1

Javascript

function merge(a) {
    var r=[], p;
    while(a.length>0) {
        for (var i=0,j=0; i<a.length && p!=a[j][0]; i++)
            if (a[i][0]<a[j][0])
                j = i;

        r.push(p = a[j].shift());

        if (!a[j].length)
            a.splice(j, 1);
    }
    return r;
}

Test:

var arr = [[1, 4, 7], [2, 5, 8], [3, 6, 9]]​;
alert(merge(arr));

Answer 2

VB is usually not the language of choice for code golf, but here goes anyway.

The setup -

        Dim m1 As List(Of Integer) = New List(Of Integer)
        Dim m2 As List(Of Integer) = New List(Of Integer)
        Dim m3 As List(Of Integer) = New List(Of Integer)
        Dim m4 As List(Of Integer) = New List(Of Integer)

        m1.Add(1)
        m1.Add(2)
        m1.Add(3)

        m2.Add(4)
        m2.Add(5)
        m2.Add(6)

        m3.Add(7)
        m3.Add(8)
        m3.Add(9)

        Dim m5 As List(Of List(Of Integer)) = New List(Of List(Of Integer))
        m5.Add(m1)
        m5.Add(m2)
        m5.Add(m3)

An attempt in VB.NET (without sort)

        While m5.Count > 0
            Dim idx As Integer = 0
            Dim min As Integer = Integer.MaxValue
            For k As Integer = 0 To m5.Count - 1
                If m5(k)(0) < min Then min = m5(k)(0) : idx = k
            Next
            m4.Add(min) : m5(idx).RemoveAt(0)
            If m5(idx).Count = 0 Then m5.RemoveAt(idx)
        End While

Another VB.NET attempt (with an allowed sort)

    Private Function Comp(ByVal l1 As List(Of Integer), ByVal l2 As List(Of Integer)) As Integer
        Return l1(0).CompareTo(l2(0))
    End Function
    .
    .
    .
    While m5.Count > 0
        m5.Sort(AddressOf Comp)
        m4.Add(m5(0)(0)) : m5(0).RemoveAt(0)
        If m5(0).Count = 0 Then m5.RemoveAt(0)
    End While

The entire program -

        Dim rand As New Random
        Dim m1 As List(Of Integer) = New List(Of Integer)
        Dim m2 As List(Of Integer) = New List(Of Integer)
        Dim m3 As List(Of Integer) = New List(Of Integer)
        Dim m4 As List(Of Integer) = New List(Of Integer)
        Dim m5 As List(Of List(Of Integer)) = New List(Of List(Of Integer))
        m5.Add(m1)
        m5.Add(m2)
        m5.Add(m3)

        For Each d As List(Of Integer) In m5
            For i As Integer = 0 To 100000
                d.Add(rand.Next())
            Next
            d.Sort()
        Next

        Dim sw As New Stopwatch
        sw.Start()
        While m5.Count > 0
            Dim idx As Integer = 0
            Dim min As Integer = Integer.MaxValue
            For k As Integer = 0 To m5.Count - 1
                If m5(k)(0) < min Then min = m5(k)(0) : idx = k
            Next
            m4.Add(min) : m5(idx).RemoveAt(0)
            If m5(idx).Count = 0 Then m5.RemoveAt(idx)
        End While
        sw.Stop()

        'Dim sw As New Stopwatch
        'sw.Start()
        'While m5.Count > 0
        '    m5.Sort(AddressOf Comp)
        '    m4.Add(m5(0)(0)) : m5(0).RemoveAt(0)
        '    If m5(0).Count = 0 Then m5.RemoveAt(0)
        'End While
        'sw.Stop()

        Console.WriteLine(sw.Elapsed)
        Console.ReadLine()

Answer 3

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!

Answer 4

F#: 116 chars:

let p l=
    let f a b=List.filter(a b) in
    let rec s=function[]->[]|x::y->s(f(>)x y)@[x]@s(f(<=)x y) in
    [for a in l->>a]|>s

Note: this code causes F# to throw a lot of warnings, but it works :)

Here's the annotated version with whitespace and meaningful identifiers (note: the code above doesn't use #light syntax, the code below does):

let golf l=
    // filters my list with a specified filter operator
    // uses built-in F# function
    // ('a -> 'b -> bool) -> 'a -> ('b list -> 'b list)
    let filter a b = List.filter(a b)

    // quicksort algorithm
    // ('a list -> 'a list)
    let rec qsort =function
        | []->[]
        | x :: y -> qsort ( filter (>) x y) @ [x] @ qsort ( filter (<=) x y)

    // flattens list
    [for a in l ->> a ] |> qsort

Answer 5

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 6

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