Sorting algorithms/Merge sort

The merge sort is a recursive sort of order n*log(n) (technically n*lg(n)--lg is log base 2) sort. The basic idea is to split the collection into smaller groups by halving it until the groups only have one element or no elements (which are both entirely sorted groups). Then merge the groups back together so that their elements are in order. This is how the algorithm gets is "divide and conquer" description.

Sorting algorithms/Merge sort
You are encouraged to solve this task according to the task description, using any language you may know.

Sorting Algorithm
This is a sorting algorithm.   It may be applied to a set of data in order to sort it.     For comparing various sorts, see compare sorts.   For other sorting algorithms,   see sorting algorithms,   or:

O(n logn) sorts

Heap sort | Merge sort | Patience sort | Quick sort

O(n log²n) sorts
Shell Sort

O(n²) sorts
Bubble sort | Cocktail sort | Cocktail sort with shifting bounds | Comb sort | Cycle sort | Gnome sort | Insertion sort | Selection sort | Strand sort

other sorts
Bead sort | Bogo sort | Common sorted list | Composite structures sort | Custom comparator sort | Counting sort | Disjoint sublist sort | External sort | Jort sort | Lexicographical sort | Natural sorting | Order by pair comparisons | Order disjoint list items | Order two numerical lists | Object identifier (OID) sort | Pancake sort | Quickselect | Permutation sort | Radix sort | Ranking methods | Remove duplicate elements | Sleep sort | Stooge sort | [Sort letters of a string] | Three variable sort | Topological sort | Tree sort

Write a function to sort a collection of integers using the merge sort. The merge sort algorithm comes in two parts: a sort function and a merge function. The functions in pseudocode look like this:

function mergesort(m)
   var list left, right, result
   if length(m) ≤ 1
       return m
   else
       var middle = length(m) / 2
       for each x in m up to middle - 1
           add x to left
       for each x in m at and after middle
           add x to right
       left = mergesort(left)
       right = mergesort(right)
       result = merge(left, right)
       return result

function merge(left,right)
   var list result
   while length(left) > 0 and length(right) > 0
       if first(left) ≤ first(right)   comment: change the direction of this comparison to change the direction of the sort
           append first(left) to result
           left = rest(left)
       else
           append first(right) to result
           right = rest(right)
   if length(left) > 0 
       append rest(left) to result
   if length(right) > 0 
       append rest(right) to result
   return result

Java

import java.util.List;
import java.util.LinkedList;
public class Merge{
	public static LinkedList<Integer> mergeSort(LinkedList<Integer> m){
		LinkedList<Integer> left = new LinkedList<Integer>();
		LinkedList<Integer> right = new LinkedList<Integer>();
		LinkedList<Integer> result = new LinkedList<Integer>();

		if (m.size() <= 1) return m;

		int middle = m.size() / 2;
		for(int i = 0; i < middle; i++) left.add(m.get(i));
		for(int i = middle; i < m.size(); i++) right.add(m.get(i));

		right = mergeSort(right);
		left = mergeSort(left);
		result = merge(left, right);

		return result;
	}

	public static LinkedList<Integer> merge(LinkedList<Integer> left, LinkedList<Integer> right){
  		LinkedList<Integer> result = new LinkedList<Integer>();

  		while (left.size() > 0 && right.size() > 0){
			//change the direction of this comparison to change the direction of the sort
			if (left.peek() <= right.peek()) result.add(left.remove());
			else result.add(right.remove());
		}

		if (left.size() > 0) result.addAll(left);
		if (right.size() > 0) result.addAll(right);
		return result;
	}
}