Merge to sorted arrays into second array, both arrays has N elements, but second array size is N * 2. So you merge both arrays in second array in a sorted way.

Question

Anónimo · Accepted Answer

// A and B are sorted input vectors and C is merged output vector
void vectorMerge(const vector &A, const vector &B, vector &C)
{
	int i = 0;
	int j = 0;
	int k = 0;

	while ((j < B.size()) && (i < A.size()))
	{
		if (A[i] <= B[j])
		{
			C[k++] = A[i++];
		}
		else
		{
			C[k++] = B[j++];
		}
	}

	if (j < B.size())
	{
		for (; j < B.size(); ++j)
		{
			C[k++] = B[j];
		}
	}

	if (i < A.size())
	{
		for (; i < A.size(); ++i)
		{
			C[k++] = A[i];
		}
	}
}

Yulia · Answer

// Example program
#include 
#include 
#include 

int main()
{
    std::vector A = {0, 2};
    std::vector B = {3, 5, 8};
    size_t size = A.size() + B.size();
    std::vector C(size);

    size_t i = 0, j = 0, k = 0;
    
    while (k < size)
    {
        if (i < A.size() && A[i] < B[j]) {
            C[k++] = A[i++];
             std::cout << C[k-1] << " ";
             continue;
        }
        
        C[k++] = B[j++];
        std::cout << C[k-1] << " ";
    }
 
}

Seedy · Answer

# merge from backwards, so that the sorted element are copied into the larger array

def mergeLists (li1, len1, li2, len2):
    idx1 = len1 - 1 
    idx2 = len2 - 1 
    idx3= len1  + len2 -1
    for i in range (idx3,-1, -1):
        if idx1 >= 0:
            val1 = li1[idx1]
        else:
            return

        if idx2 >= 0:
            val2 = li2[idx2]
        else:
            for j in range (idx1+1):
                li2[j] = li1[j]
            return

        if val1 >= val2:
            li2[i] = val1
            idx1 -= 1 
        else:
            li2[i] =  val2
            idx2 -= 1 

li1 = [1,4,9, 15] 
li2 = [3,6,9,10,19,0,0,0,0]
print (li1, li2)
mergeLists (li1,4, li2, 5)
print (li1, li2)

Anónimo · Answer

In a regular merge sort, the arrays are sorted from the beginning. Here, sort from the end directly to the second array.

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