C++ Merge Sort using iterators - c++

Question

Currently, I am trying to create a simple C++ Merge Sort Program.
using namespace std;
using Iterator = std::vector<int>::iterator;
using CIterator = std::vector<int>::const_iterator;
std::vector<int> merge(CIterator left_begin, CIterator left_end, CIterator right_begin, CIterator right_end) {
std::vector<int> result;
CIterator left = left_begin;
CIterator right = right_begin;
while (left != left_end && right != right_end) {
if (*left <= *right) {
result.push_back(*left);
left++;
} else {
result.push_back(*right);
right++;
}
}
while (left != left_end) {
result.push_back(*left);
left++;
}
while (right != right_end) {
result.push_back(*right);
right++;
}
return result;
}
I created a merge function that basically connects two sorted vectors into one and returns it (I am bound to use the following return type of the function merge). Then Trying to write the driver function merge sort I have the following code, that I think works correctly
void merge_sort(Iterator begin, Iterator end) {
auto difference = distance(begin, end);
if (difference <= 1) {
return;
}
Iterator middle = begin;
advance(middle, difference / 2);
merge_sort(begin, middle);
merge_sort(middle, end);
vector<int> result = merge(begin, middle, middle, end);
// But what to put here?
}
At the place of the comment mark, I don't understand what to write in order to move the sorted array a step up in the recursion. I tried
begin = result.begin();
end = result.end();
but this obviously doesnt work

Answer 1

The problem is that the type signature for merge_sort assumes an in-place algorithm:
void merge_sort(Iterator begin, Iterator end);
But your merge procedure isn't in-place but returns a merged copy of the arrays. You either need to change merge to be in-place, or you need to change merge_sort to return the sorted array. The latter solution (easier but less efficient) would be like this:
std::vector<int> merge_sort(Iterator begin, Iterator end) {
auto difference = distance(begin, end);
if (difference <= 1) {
return;
}
Iterator middle = begin;
advance(middle, difference / 2);
std::vector<int> left = merge_sort(begin, middle);
std::vector<int> right = merge_sort(middle, end);
return merge(left.begin(), left.end(), right.begin(), right.end());
}

Answer 2

An optimized top down merge sort that does a one time allocation of a second vector, then uses a pair of mutually recursive functions (...AtoA, ...AtoB) to alternate the direction of merge based on level of recursion. (I left out the prototypes).
void MergeSort( typename std::vector<int>::iterator &ab,
typename std::vector<int>::iterator &ae)
{
size_t sz = ae - ab;
if (sz < 2)
return;
std::vector<int> vb(sz); // temp vector
std::vector<int>::iterator bb = vb.begin();
std::vector<int>::iterator be = vb.end();
MergeSortAtoA(ab, ae, bb, be);
}
void MergeSortAtoA( typename std::vector<int>::iterator &ab,
typename std::vector<int>::iterator &ae,
typename std::vector<int>::iterator &bb,
typename std::vector<int>::iterator &be)
{
size_t sz = ae - ab;
if(sz < 2) // if 1 element return
return;
std::vector<int>::iterator am = ab+(sz/2);
std::vector<int>::iterator bm = bb+(sz/2);
MergeSortAtoB(ab, am, bb, bm);
MergeSortAtoB(am, ae, bm, be);
Merge(bb, bm, be, ab);
}
void MergeSortAtoB( typename std::vector<int>::iterator &ab,
typename std::vector<int>::iterator &ae,
typename std::vector<int>::iterator &bb,
typename std::vector<int>::iterator &be)
{
size_t sz = ae - ab;
if(sz < 2){ // if 1 element, copy it
*bb = *ab;
return;
}
std::vector<int>::iterator am = ab+(sz/2);
std::vector<int>::iterator bm = bb+(sz/2);
MergeSortAtoA(ab, am, bb, bm);
MergeSortAtoA(am, ae, bm, be);
Merge(ab, am, ae, bb);
}
void Merge( typename std::vector<int>::iterator &ab,
typename std::vector<int>::iterator &am,
typename std::vector<int>::iterator &ae,
typename std::vector<int>::iterator &bb)
{
std::vector<int>::iterator mb = ab; // left run iterator
std::vector<int>::iterator mm = am; // right run iterator
std::vector<int>::iterator bi = bb; // merge run iterator
while(1){ // merge data
if(*mb <= *mm){ // if mb <= mm
*bi++ = *mb++; // copy mb
if(mb < am) // if not end left run
continue; // continue (back to while)
while(mm < ae) // else copy rest of right run
*bi++ = *mm++;
break; // and return
} else { // else mb > mm
*bi++ = *mm++; // copy mm
if(mm < ae) // if not end of right run
continue; // continue (back to while)
while(mb < am) // else copy rest of left run
*bi++ = *mb++;
break; // and return
}
}
}