Algorithm Implementation/Sorting/Merge sort
Merge Sort
editYou start with an unordered sequence. You create N empty queues. You loop over every item to be sorted. On each loop iteration, you look at the last element in the key. You move that item into the end of the queue which corresponds to that element. When you are finished looping you concatenate all the queues together into another sequence. You then reapply the procedure described but look at the second last element in the key. You keep doing this until you have looped over every key. When you complete this process the resulting sequence will be sorted as described above.
Key Comparing
editKeys are compared in the following way: Let ka be the key of the one item, called item A, let kb be the key of the other item, called item B. Let ka(i) be the ith entry in the key ka, where the first entry is at index 0. Let i = 0. If the keys are less than i elements long then the keys are equal. If ka(i) < kb(i), then item A is ordered before item B. If ka(i) > kb(i), then item B is ordered before item A. If ka(i) = kb(i), then add one to i, and return the line under "Let i = 0."
Time Cost
editLet ni be the number of items in the sequence to be sorted. N is number of integers that each key element can take. Let nk be the number of keys in each item.
The total time to sort the sequence is thus O(nk(ni + N)).
Naive implementation, translation of pseudocode found at Wikipedia.
(defmacro
apenda-primeiro (ret1 left1)
"Appends first element of left1 to right1, and removes first element from left1."
`(progn
(setf ,ret1 (if (eq ,ret1 nil) (cons (car ,left1) nil) (append ,ret1 (cons (car ,left1) nil))))
(setf ,left1 (cdr ,left1))))
(defun
merge-func (left right)
"Our very own merge function, takes two lists, left and right, as arguments, and returns a new merged list."
(let (ret)
(loop
(if (or (not (null left)) (not (null right)))
(progn
(if (and (not (null left)) (not (null right)))
(if (<= (car left) (car right))
(apenda-primeiro ret left)
(apenda-primeiro ret right))
(if (not (null left))
(apenda-primeiro ret left)
(if (not (null right))
(apenda-primeiro ret right)))))
(return)))
ret))
(defun
merge-sort (m)
"Merge-sort proper. Takes a list m as input and returns a new, sorted list; doesn't touch the input."
(let* ((tam (length m))
(mid (ceiling (/ tam 2)))
(left)
(right))
(if (<= tam 1)
m
(progn
(loop for n from 0 to (- mid 1) do
(apenda-primeiro left m))
(loop for n from mid to (- tam 1) do
(apenda-primeiro right m))
(setf left (merge-sort left))
(setf right (merge-sort right))
(merge-func left right)))))
Simpler Implementation in a somewhat more functional style.
(defun sort-func (frst scond &optional (sorted nil))
"Sorts the elements from the first and second list in ascending order and puts them in `sorted`"
(cond ((and (null frst) (null scond)) sorted)
((null frst) (append (reverse sorted) scond))
((null scond) (append (reverse sorted) frst))
(t (let ((x (first frst))
(y (first scond)))
(if (< x y)
(sort-func (rest frst) scond (push x sorted))
(sort-func frst (rest scond) (push y sorted)))))))
(defun merge-sort (lst)
"Divides the elements in `lst` into individual elements and sorts them"
(when (not (null lst))
(let ((divided (mapcar #'(lambda (x) (list x)) lst)))
(labels ((merge-func (div-list &optional (combined '())) ; merges the elements in ascending order
(if div-list
(merge-func (rest (rest div-list)) (push (sort-func (first div-list) (second div-list)) combined))
combined))
(final-merge (div-list) ; runs merge-func until all elements have been reconciled
(if (> (length div-list) 1)
(final-merge (merge-func div-list))
div-list)))
(final-merge divided)))))
///function:
mergeSort(name_array);
//tipo Data used:
typedef struct data{
char*some;
int data;
} DATA;
typedef struct _nodo{
int key;
DATA data;
}nodo;
///n is kept as global
int n;
void merge(nodo*a,nodo*aux,int left,int right,int rightEnd){
int i,num,temp,leftEnd=right-1;
temp=left;
num=rightEnd-left+1;
while((left<=leftEnd)&&(right<=rightEnd)){
if(a[left].key<=a[right].key){
aux[temp++]=a[left++];
}
else{
aux[temp++]=a[right++];
}
}
while(left<=leftEnd){
aux[temp++]=a[left++];
}
while(right<=rightEnd){
aux[temp++]=a[right++];
}
for (i=1;i<=num;i++,rightEnd--){
a[rightEnd]=aux[rightEnd];
}
}
void mSort(nodo*a,nodo*aux,int left,int right){
int center;
if (left<right){
center=(left+right)/2;
mSort(a,aux,left,center);
mSort(a,aux,center+1,right);
merge(a,aux,left,center+1,right);
}
}
void mergeSort(nodo*p){
nodo*temp=(nodo*)malloc(sizeof(nodo)*n);
mSort(p,temp,0,n-1);
free(temp);
}
A recursive implementation using the C++14 standard library.
#include <iterator>
#include <algorithm>
#include <functional>
template <typename BidirIt, typename Compare = std::less<>>
void merge_sort(BidirIt first, BidirIt last, Compare cmp = Compare {})
{
const auto n = std::distance(first, last);
if (n > 1) {
const auto middle = std::next(first, n / 2);
merge_sort(first, middle, cmp);
merge_sort(middle, last, cmp);
std::inplace_merge(first, middle, last, cmp);
}
}
#include <vector>
int main()
{
std::vector<int> v {3, -2, 1, 5, -9, 10, 3, -3, 2};
merge_sort(std::begin(v), std::end(v)); // sort increasing
merge_sort(std::begin(v), std::end(v), std::greater<> {}); // sort decreasing
}
subroutine Merge(A,NA,B,NB,C,NC)
integer, intent(in) :: NA,NB,NC ! Normal usage: NA+NB = NC
integer, intent(in out) :: A(NA) ! B overlays C(NA+1:NC)
integer, intent(in) :: B(NB)
integer, intent(in out) :: C(NC)
integer :: I,J,K
I = 1; J = 1; K = 1;
do while(I <= NA .and. J <= NB)
if (A(I) <= B(J)) then
C(K) = A(I)
I = I+1
else
C(K) = B(J)
J = J+1
endif
K = K + 1
enddo
do while (I <= NA)
C(K) = A(I)
I = I + 1
K = K + 1
enddo
return
end subroutine merge
recursive subroutine MergeSort(A,N,T)
integer, intent(in) :: N
integer, dimension(N), intent(in out) :: A
integer, dimension((N+1)/2), intent (out) :: T
integer :: NA,NB,V
if (N < 2) return
if (N == 2) then
if (A(1) > A(2)) then
V = A(1)
A(1) = A(2)
A(2) = V
endif
return
endif
NA=(N+1)/2
NB=N-NA
call MergeSort(A,NA,T)
call MergeSort(A(NA+1),NB,T)
if (A(NA) > A(NA+1)) then
T(1:NA)=A(1:NA)
call Merge(T,NA,A(NA+1),NB,A,N)
endif
return
end subroutine MergeSort
program TestMergeSort
integer, parameter :: N = 8
integer, dimension(N) :: A = (/ 1, 5, 2, 7, 3, 9, 4, 6 /)
integer, dimension ((N+1)/2) :: T
call MergeSort(A,N,T)
write(*,'(A,/,10I3)')'Sorted array :',A
end program TestMergeSort
function merge_sort(arr)
if length(arr) <= 1
return arr
end
middle = trunc(Int, length(arr) / 2)
L = arr[1:middle]
R = arr[middle+1:end]
merge_sort(L)
merge_sort(R)
i = j = k = 1
while i <= length(L) && j <= length(R)
if L[i] < R[j]
arr[k] = L[i]
i+=1
else
arr[k] = R[j]
j+=1
end
k+=1
end
while i <= length(L)
arr[k] = L[i]
i+=1
k+=1
end
while j <= length(R)
arr[k] = R[j]
j+=1
k+=1
end
arr
end
(define (mergesort x)
(if (= 0 (length x))
'()
;else
(if (= (length x) 1)
x
;else
(combine (mergesort (firstHalf x (/ (length x) 2))) (mergesort (lastHalf x (/ (length x) 2)))
)
)
)
)
(define (combine list1 list2)
(if (null? list1) list2
;else
(if (null? list2) list1
;else
(if (<= (car list1) (car list2))
;car of list 1 is second element of list 2
(cons (car list1) (combine (cdr list1) list2))
;else
(cons (car list2) (combine list1 (cdr list2)))
)
)
)
)
(define (firstHalf L N)
(if (= N 0)
null
)
(if (or (= N 1) (< N 2))
(list (car L))
;else
(cons (car L) (firstHalf (cdr L) (- N 1)))))
(define (lastHalf L N)
(if (= N 0) L); Base Case
(if (or (= N 1) (< N 2))
(cdr L)
;else
(lastHalf (cdr L) (- N 1)))
)
sort :: Ord a => [a] -> [a]
sort [] = []
sort [x] = [x]
sort xs = merge (sort ys) (sort zs)
where
(ys,zs) = splitAt (length xs `div` 2) xs
merge [] y=y
merge x []=x
merge (x:xs) (y:ys)
| x<=y = x:merge xs (y:ys)
| otherwise = y:merge (x:xs) ys
A slightly more efficient version only traverses the input list once to split (note that length takes linear time in Haskell):
sort :: Ord a => [a] -> [a]
sort [] = []
sort [x] = [x]
sort xs = merge (sort ys) (sort zs)
where (ys,zs) = split xs
merge [] ys = ys
merge xs [] = xs
merge (x:xs) (y:ys)
| x<=y = x : merge xs (y:ys)
| otherwise = y : merge (x:xs) ys
split [] = ([], [])
split [x] = ([x], [])
split (x:y:xs) = (x:l, y:r)
where (l, r) = split xs
function merge(sequence left, sequence right)
sequence result = {}
while length(left)>0 and length(right)>0 do
if left[1]<=right[1] then
result = append(result, left[1])
left = left[2..$]
else
result = append(result, right[1])
right = right[2..$]
end if
end while
return result & left & right
end function
function mergesort(sequence m)
sequence left, right
integer middle
if length(m)<=1 then
return m
end if
middle = floor(length(m)/2)
left = mergesort(m[1..middle])
right = mergesort(m[middle+1..$])
if left[$]<=right[1] then
return left & right
elsif right[$]<=left[1] then
return right & left
end if
return merge(left, right)
end function
This is an ISO-Prolog compatible implementation of merge sort.
mergesort(L, Sorted) :-
once(mergesort_r(L, Sorted)).
mergesort_r([], []).
mergesort_r([X], [X]).
mergesort_r(L, Sorted) :-
split(L, Left, Right),
mergesort_r(Left, SortedL),
mergesort_r(Right, SortedR),
merge(SortedL, SortedR, Sorted).
% Split list into two roughly equal-sized lists.
split([], [], []).
split([X], [X], []).
split([X,Y|Xs], [X|Left], [Y|Right]) :-
split(Xs, Left, Right).
% Merge two sorted lists into one.
merge(Left, [], Left).
merge([], Right, Right).
merge([X|Left], [Y|Right], [Z|Merged]) :-
(X @< Y ->
Z = X,
merge(Left, [Y|Right], Merged)
;
Z = Y,
merge([X|Left], Right, Merged)
).
A "standard" mergesort:
from heapq import merge
def mergesort(w):
"""Sort list w and return it."""
if len(w)<2:
return w
else:
# sort the two halves of list w recursively with mergesort and merge them
return merge(mergesort(w[:len(w)//2]), mergesort(w[len(w)//2:]))
An alternative method, using a recursive algorithm to perform the merging in place (except for the O(log n) overhead to trace the recursion) in O(n log n) time:
def merge(lst, frm, pivot, to, len1, len2):
if len1==0 or len2==0: return
if len1+len2 == 2:
if lst([pivot])<lst[frm]: lst[pivot], lst[frm] = lst[frm], lst[pivot]
return
if len1 > len2:
len11 = len1/2
firstcut, secondcut, length = frm+len11, pivot, to-pivot
while length > 0:
half = length/2
mid = secondcut+half
if lst[mid]<lst[firstcut]: secondcut, length = mid+1, length-half-1
else: length = half
len22 = secondcut - pivot
else:
len22 = len2/2
firstcut, secondcut, length = frm, pivot+len22, pivot-frm
while length > 0:
half = length/2
mid = firstcut+half
if lst[secondcut]<lst[mid]: length = half
else: firstcut, length = mid+1, length-half-1
len11 = firstcut-frm
if firstcut!=pivot and pivot!=secondcut:
n, m = secondcut-firstcut, pivot-firstcut
while m != 0: n, m = m, n%m
while n != 0:
n -= 1
p1, p2 = firstcut+n, n+pivot
val, shift = lst[p1], p2-p1
while p2 != firstcut+n:
lst[p1], p1 = lst[p2], p2
if secondcut-p2>shift: p2 += shift
else: p2 = pivot-secondcut+p2
lst[p1] = val
newmid = firstcut+len22
merge(lst, frm, firstcut, newmid, len11, len22)
merge(lst, newmid, secondcut, to, len1-len11, len2-len22)
def sort(lst, frm=0, to=None):
if to is None: to = len(lst)
if to-frm<2: return
middle = (frm+to)/2
sort(lst, frm, middle)
sort(lst, middle, to)
merge(lst, frm, middle, to, middle-frm, to-middle)
def merge_sort(array)
return array if array.size <= 1
mid = array.size / 2
merge(merge_sort(array[0...mid]), merge_sort(array[mid...array.size]))
end
def merge(left, right)
result = []
until left.empty? || right.empty?
result << (left[0] <= right[0] ? left : right).shift
end
result.concat(left).concat(right)
end
sort [] = []
sort [x] = [x]
sort array = merge (sort left) (sort right)
where
left = [array!y | y <- [0..mid]]
right = [array!y | y <- [(mid+1)..max]]
max = #array - 1
mid = max div 2
fun mergesort [] = []
| mergesort [x] = [x]
| mergesort lst =
let fun merge ([],ys) = ys (*merges two sorted lists to form a sorted list *)
| merge (xs,[]) = xs
| merge (x::xs,y::ys) =
if x<y then
x::merge (xs,y::ys)
else
y::merge (x::xs,ys)
;
val half = length(lst) div 2;
in
merge (mergesort (List.take (lst, half)),mergesort (List.drop (lst, half)))
end
;
public int[] mergeSort(int array[])
// pre: array is full, all elements are valid integers (not null)
// post: array is sorted in ascending order (lowest to highest)
{
// if the array has more than 1 element, we need to split it and merge the sorted halves
if(array.length > 1)
{
// number of elements in sub-array 1
// if odd, sub-array 1 has the smaller half of the elements
// e.g. if 7 elements total, sub-array 1 will have 3, and sub-array 2 will have 4
int elementsInA1 = array.length / 2;
// we initialize the length of sub-array 2 to
// equal the total length minus the length of sub-array 1
int elementsInA2 = array.length - elementsInA1;
// declare and initialize the two arrays once we've determined their sizes
int arr1[] = new int[elementsInA1];
int arr2[] = new int[elementsInA2];
// copy the first part of 'array' into 'arr1', causing arr1 to become full
for(int i = 0; i < elementsInA1; i++)
arr1[i] = array[i];
// copy the remaining elements of 'array' into 'arr2', causing arr2 to become full
for(int i = elementsInA1; i < elementsInA1 + elementsInA2; i++)
arr2[i - elementsInA1] = array[i];
// recursively call mergeSort on each of the two sub-arrays that we've just created
// note: when mergeSort returns, arr1 and arr2 will both be sorted!
// it's not magic, the merging is done below, that's how mergesort works :)
arr1 = mergeSort(arr1);
arr2 = mergeSort(arr2);
// the three variables below are indexes that we'll need for merging
// [i] stores the index of the main array. it will be used to let us
// know where to place the smallest element from the two sub-arrays.
// [j] stores the index of which element from arr1 is currently being compared
// [k] stores the index of which element from arr2 is currently being compared
int i = 0, j = 0, k = 0;
// the below loop will run until one of the sub-arrays becomes empty
// in my implementation, it means until the index equals the length of the sub-array
while(arr1.length != j && arr2.length != k)
{
// if the current element of arr1 is less than current element of arr2
if(arr1[j] < arr2[k])
{
// copy the current element of arr1 into the final array
array[i] = arr1[j];
// increase the index of the final array to avoid replacing the element
// which we've just added
i++;
// increase the index of arr1 to avoid comparing the element
// which we've just added
j++;
}
// if the current element of arr2 is less than current element of arr1
else
{
// copy the current element of arr2 into the final array
array[i] = arr2[k];
// increase the index of the final array to avoid replacing the element
// which we've just added
i++;
// increase the index of arr2 to avoid comparing the element
// which we've just added
k++;
}
}
// at this point, one of the sub-arrays has been exhausted and there are no more
// elements in it to compare. this means that all the elements in the remaining
// array are the highest (and sorted), so it's safe to copy them all into the
// final array.
while(arr1.length != j)
{
array[i] = arr1[j];
i++;
j++;
}
while(arr2.length != k)
{
array[i] = arr2[k];
i++;
k++;
}
}
// return the sorted array to the caller of the function
return array;
}
;(function() {
var defaultComparator = function (a, b) {
if (a < b) {
return -1;
}
if (a > b) {
return 1;
}
return 0;
}
Array.prototype.mergeSort = function( comparator ) {
var i, j, k,
firstHalf,
secondHalf,
arr1,
arr2;
if (typeof comparator != "function") { comparator = defaultComparator; }
if (this.length > 1) {
firstHalf = Math.floor(this.length / 2);
secondHalf = this.length - firstHalf;
arr1 = [];
arr2 = [];
for (i = 0; i < firstHalf; i++) {
arr1[i] = this[i];
}
for(i = firstHalf; i < firstHalf + secondHalf; i++) {
arr2[i - firstHalf] = this[i];
}
arr1.mergeSort( comparator );
arr2.mergeSort( comparator );
i=j=k=0;
while(arr1.length != j && arr2.length != k) {
if ( comparator( arr1[j], arr2[k] ) <= 0 ) {
this[i] = arr1[j];
i++;
j++;
}
else {
this[i] = arr2[k];
i++;
k++;
}
}
while (arr1.length != j) {
this[i] = arr1[j];
i++;
j++;
}
while (arr2.length != k) {
this[i] = arr2[k];
i++;
k++;
}
}
}
})();
Separate into two functions:
function mergesort(list){
return (list.length < 2) ? list : merge(mergesort(list.splice(0, list.length >> 1)), mergesort(list));
}
function merge(left, right){
var sorted = [];
while (left.length && right.length)
sorted.push(left[0] <= right[0]? left.shift() : right.shift());
while(left.length)
sorted.push(left.shift());
while(right.length)
sorted.push(right.shift());
return sorted;
}
use sort '_mergesort';
sort @array;
function merge_sort(array $left, array $right) {
$result = [];
while (count($left) && count($right))
($left[0] < $right[0]) ? $result[] = array_shift($left) : $result[] = array_shift($right);
return array_merge($result, $left, $right);
}
function merge(array $arrayToSort) {
if (count($arrayToSort) == 1)
return $arrayToSort;
$left = merge(array_slice($arrayToSort, 0, count($arrayToSort) / 2));
$right = merge(array_slice($arrayToSort, count($arrayToSort) / 2, count($arrayToSort)));
return merge_sort($left, $right);
}
def mergeSort(def list){
if (list.size() <= 1) { return list }
else {
center = list.size() / 2
left = list[0..center]
right = list[center..list.size() - 1]
merge(mergeSort(left), mergeSort(right))
}
}
def merge(def left, def right){
def sorted = []
while(left.size() > 0 && right.size() > 0)
if(left.get(0) <= right.get(0)){
sorted << left.remove(0)
}else{
sorted << right.remove(0)
}
sorted = sorted + left + right
return sorted
}
class
APPLICATION
feature -- Algorithm
mergesort (a: ARRAY [INTEGER]; l, r: INTEGER)
-- Recursive mergesort
local
m: INTEGER
do
if l < r then
m := (l + r) // 2
mergesort(a, l, m)
mergesort(a, m + 1, r)
merge(a, l, m, r)
end
end
feature -- Utility feature
merge (a: ARRAY [INTEGER]; l, m, r: INTEGER)
-- The merge feature of all mergesort variants
local
b: ARRAY [INTEGER]
h, i, j, k: INTEGER
do
i := l
j := m + 1
k := l
create b.make (l, r)
from
until
i > m or j > r
loop
-- Begins the merge and copies it into an array `b'
if a.item (i) <= a.item (j) then
b.item (k) := a.item (i)
i := i + 1
elseif a.item (i) > a.item (j) then
b.item (k) := a.item (j)
j := j + 1
end
k := k + 1
end
-- Finishes the copy of the uncopied part of the array
if i > m then
from
h := j
until
h > r
loop
b.item (k + h - j) := a.item (h)
h := h + 1
end
elseif j > m then
from
h := i
until
h > m
loop
b.item (k + h - i) := a.item (h)
h := h + 1
end
end
-- Begins the copy to the real array
from
h := l
until
h > r
loop
a.item (h) := b.item (h)
h := h + 1
end
end
feature -- Attributes
array: ARRAY [INTEGER]
end
C#
editpublic class MergeSort<T> where T : IComparable
{
public T[] Sort(T[] source)
{
T[] sorted = this.Split(source);
return sorted;
}
private T[] Split(T[] from)
{
if (from.Length == 1)
{
// size <= 1 considered sorted
return from;
}
else
{
int iMiddle = from.Length / 2;
T[] left = new T[iMiddle];
for (int i = 0; i < iMiddle; i++)
{
left[i] = from[i];
}
T[] right = new T[from.Length - iMiddle];
for (int i = iMiddle; i < from.Length; i++)
{
right[i - iMiddle] = from[i];
}
// Single threaded version
T[] sLeft = this.Split(left);
T[] sRight = this.Split(right);
T[] sorted = this.Merge(sLeft, sRight);
return sorted;
}
}
private T[] Merge(T[] left, T[] right)
{
// each array will individually be sorted.
// Do a sort of card merge to merge them in a sorted sequence
int leftLen = left.Length;
int rightLen = right.Length;
T[] sorted = new T[leftLen + rightLen];
int lIndex = 0;
int rIndex = 0;
int currentIndex = 0;
while (lIndex < leftLen || rIndex < rightLen)
{
// If either has had all elements taken, just take remaining from the other.
// If not, compare the two current and take the lower.
if (lIndex >= leftLen)
{
sorted[currentIndex] = right[rIndex];
rIndex++;
currentIndex++;
}
else if (rIndex >= rightLen)
{
sorted[currentIndex] = left[lIndex];
lIndex++;
currentIndex++;
}
else if (left[lIndex].CompareTo(right[rIndex]) >= 0)
{
// l > r, so r goes into dest
sorted[currentIndex] = right[rIndex];
rIndex++;
currentIndex++;
}
else
{
sorted[currentIndex] = left[lIndex];
lIndex++;
currentIndex++;
}
}
return sorted;
}
}
const maxA = 1000;
type TElem = integer;
TArray = array[1..maxA]of TElem;
procedure merge(var A:TArray;p,q,r:integer);
var i,j,k,n1,n2:integer;
B:TArray;
begin
n1 := q - p + 1;
n2 := r - q;
for k := p to r do
B[k - p + 1] := A[k];
i := 1;
j :=n1 + 1;
k := p;
while(i <= n1)and(j <= n1 + n2)do
begin
if B[i] <= B[j] then
begin
A[k] := B[i];
i := i + 1;
end
else
begin
A[k] := B[j];
j := j + 1;
end;
k := k + 1;
end;
while i <= n1 do
begin
A[k] := B[i];
i := i + 1;
k := k + 1;
end;
while j <= n1 + n2 do
begin
A[k] := B[j];
j := j + 1;
k := k + 1;
end;
end;
(* Recursive version of merge sort *)
procedure mergeSort(var A:TArray;p,r:integer);
var q:integer;
begin
if p < r then
begin
q := (p + r) div 2;
mergeSort(A,p,q);
mergeSort(A,q + 1,r);
merge(A,p,q,r);
end;
end;
(* Iterative version of merge sort *)
procedure mergeSort(var A:TArray;n:integer);
var p,q,r,k:integer;
begin
k := 1;
while k <= n do
begin
p := 1;
while p + k <= n do
begin
q := p + k - 1;
if p + 2 * k - 1 < n then
r := p + 2 * k - 1
else
r := n;
merge(A,p,q,r);
p := p + 2 * k;
end;
k := k * 2;
end;
end;
Using a functor to create modules that specialize sorting lists of a given type with a particular comparison function:
module type Comparable = sig
type t
val compare: t -> t -> int
end
module MakeSorter(M : Comparable) = struct
(** Split list into two roughly equal halves *)
let partition (lst: M.t list) =
let rec helper lst left right =
match lst with
| [ ] -> left, right
| x :: [ ] -> x :: left, right
| x :: y :: xs -> helper xs (x :: left) (y :: right) in
helper lst [] []
(** Merge two sorted lists *)
let rec merge left right =
match left, right with
| _, [ ] -> left (* Emmpty right list *)
| [ ], _ -> right (* Empty left list *)
| x :: xs, y :: _ when (M.compare x y) < 0 -> x :: merge xs right (* First element of left is less than first element of right *)
| _, y :: ys -> y :: merge left ys (* First element of right is greater than or equal to first element of left *)
let rec sort lst =
match lst with
| [ ] | _ :: [ ] -> lst (* Empty and single element lists are always sorted *)
| lst ->
let left, right = partition lst in
merge (sort left) (sort right)
end
module StringSort = MakeSorter(String)
let () =
let animals = [ "dog"; "cow"; "ant"; "zebra"; "parrot" ] in
let sorted_animals = StringSort.sort animals in
List.iter print_endline sorted_animals