Split given list into two sublists

Given a list, split it into two sublists — one for the front half, and one for the back half. If the number of elements is odd, the extra element should go in the front list. So FrontBackSplit() on the list {2, 3, 5, 7, 11} should yield the two lists {2, 3, 5} and {7,11}.

Trickier way to do this problem is to have two pointers which traverse the list. One pointer advances two nodes at a time, while the other goes one node at a time. When the leading pointer reaches the end, the trailing pointer will be about half way.

/*
 * Split the nodes of the given list into front and back halves,
 * and return the two lists using the reference parameters.
 * If the length is odd, the extra node should go in the front list.
 */
void FrontBackSplit(nodeptr source, nodeptr* frontRef, nodeptr* backRef)
{
    nodeptr singlestep = NULL;
    nodeptr doublestep = NULL;
    if(source == NULL) {
        *frontRef = NULL;
        *backRef = NULL;
    } else {
        *frontRef = source;
        doublestep = source->next;
        singlestep = source;
        while(doublestep != NULL) {
            if(doublestep->next != NULL)
                doublestep = doublestep->next->next;
            else 
                break;
            singlestep = singlestep->next;
        }
        *backRef = singlestep->next;
        singlestep->next = NULL;
    }
}