两有序链表合并排序

1、已知两个链表head1 和head2 各自有序，请把它们合并成一个链表依然有序。(保留所有结点，即便大小相同）

Node * Merge(Node *head1 , Node *head2)
{
    if ( head1 == NULL)
        return head2 ;
    if ( head2 == NULL)
        return head1 ;
    Node *head = NULL ;
    Node *p1 = NULL;
    Node *p2 = NULL;
    if ( head1->data < head2->data )
    {
        head = head1 ;
        p1 = head1->next;
        p2 = head2 ;
    }
    else
    {
        head = head2 ;
        p2 = head2->next ;
        p1 = head1 ;
    }
    Node *pcurrent = head ;
    while ( p1 != NULL && p2 != NULL)
    {
        if ( p1->data <= p2->data )
        {
            pcurrent->next = p1 ;
            pcurrent = p1 ;
            p1 = p1->next ;
        }
        else
        {
            pcurrent->next = p2 ;
            pcurrent = p2 ;
            p2 = p2->next ;
        }
    }
    if ( p1 != NULL )
    pcurrent->next = p1 ;
    if ( p2 != NULL )
    pcurrent->next = p2 ;
    return head ;
}

2、已知两个链表head1 和head2 各自有序，请把它们合并成一个链表依然有序，这次要求用递归方法进行。

Node * MergeRecursive(Node *head1 , Node *head2)
{
    if ( head1 == NULL )
        return head2 ;
    if ( head2 == NULL)
        return head1 ;
    Node *head = NULL ;
    if ( head1->data < head2->data )
    {
        head = head1 ;
        head->next = MergeRecursive(head1->next,head2);
    }
    else
    {
        head = head2 ;
        head->next = MergeRecursive(head1,head2->next);
    }
    return head ;
}