Merge two sorted arrays together

Question

Please review my answer for this interview question:

#include <iostream>
#include <vector>
std::vector<int> merge2Sorted ( std::vector<int> left, std::vector<int> right )
{
 //finger matching algo
 auto itLeft = left.begin();
 auto itRight = right.begin();
 auto itLeftEnd = left.end();
 auto itRightEnd = right.end();
 std::vector<int> result;
 result.reserve(std::max(left.size(),right.size())); 
 while ( itLeft != itLeftEnd && 
 itRight != itRightEnd )
 {
 if ( *itLeft < *itRight )
 result.push_back( *itLeft++ );
 else
 result.push_back( *itRight++ );
 }
 // copy rest of left array
 while ( itLeft != itLeftEnd )
 result.push_back(*itLeft++);
 // copy rest of right array
 while ( itRight != itRightEnd )
 result.push_back(*itRight++);
 return result;
}
int main ()
{
 std::vector<int> v1 = { 1,2,3,4,5,6,7,8,9,10 };
 std::vector<int> v2 = { -1,-2,0,3,7,9,11,12 };
 std::vector<int> v3 ( merge2Sorted ( v1, v2 ) );
 for ( auto& i: v3 )
 std::cout << i << std::endl;
}

Answer 1

Pass your parameters by const reference to avoid a copy:

std::vector<int> merge2Sorted(std::vector<int> const& left, std::vector<int> const& right)
// ^^^^^^ ^^^^^^

You are not reservng enough

result.reserve(std::max(left.size(),right.size()));

The result size will eventually be the size of the sum of the two input arrays.

result.reserve(left.size() + right.size());

Your test for left and right can be simplified:

if ( *itLeft < *itRight )
 result.push_back( *itLeft++ );
else
 result.push_back( *itRight++ );
}
// Simplified
// Though I am 50/50 on this one.
result.push_back( ( *itLeft < *itRight ) ? *itLeft++ : *itRight++);

No point in flushing the stream after every print:

std::cout << i << std::endl;
 ^^^^^^^^^^ prefer '\n' unless you really want to force a flush.
std::cout << i << '\n';

Answer 2

In addition to Loki's answer you can also simplify copying the rest of the left and right arrays like so:

// Copy rest of left and right vectors
result.insert(result.end(), itLeft, itLeftEnd);
result.insert(result.end(), itRight, itRightEnd);

Oh and if you were doing this in production and not for an interview you could instead use std::merge:

#include <algorithm>
std::vector<int> merge2Sorted ( const std::vector<int>& left, const std::vector<int>& right ) {
 std::vector<int> output;
 std::merge(left.begin(), left.end(), right.begin(), right.end(), std::back_inserter(output));
 return output;
}