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

Python, 107 chars:

def f(l):  
 n=[]  
 for t in l:  
  for i in t: n+=[t]  
 s=[]  
 while n: s.+=[min(n)]; n.remove(min(n))  
 return s  

Answer 2

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 3

F#, 32 chars

let f x=List.sort(List.concat x)

And without using a built in function for the concat (57 chars):

let f x=List.sort(Seq.toList(seq{for l in x do yield!l}))

Answer 4

I don't think you can get much better than @Sykora's response, here, for Python.

Changed to handle your inputs:

import heapq
def m(i): 
    return list(heapq.merge(*i))

print m(((1, 4, 7), (2, 5, 8), (3, 6, 9)))

For the actual function, 59 characters, or the 52 in reduced version:

import heapq
def m(i): return list(heapq.merge(*i))

This also has the benefit of using a tested and true implementation built into Python

Edit: Removed the semi-colons (thanks @Douglas).

Answer 5

I'll just leave this here...

Language: C, Char count: 265

L[99][99];N;n[99];m[99];i;I;b=0;main(char t){while(scanf("%d%c",L[i]+I,&t)+1){++
I;if(t==10){n[i++]=I;I=0;}}if(I)n[i++] = I;N=i;while(b+1){b=-1;for(i=0;i<N;++i){
I=m[i];if(I-n[i])if(b<0||L[i][I]<L[b][m[b]])b=i;}if(b<0)break;printf("%d ",L[b][
m[b]]);++m[b];}puts("");}

Takes input like such:

1 4 7
2 5 8
3 6 9
(EOF)

Answer 6

Even though it might break the rules. Here's a nice and short c++ entry:

13 Characters

l1.merge(l2); // Removes the elements from the argument list, inserts 
              // them into the target list, and orders the new, combined 
              // set of elements in ascending order or in some other 
              // specified order.