设计一个算法将一个带头节点的单链表A分解成两个带头结点的单链表A和B，使A中含有原链表中序号为奇数的元素，而B中含有原链表中序号为偶数的元素，并且保持元素原有的相对顺序。
这个题，我不是很清楚，请各位帮帮忙。

#include <stdio.h>
#include <stdlib.h>

typedef struct node
{
char data;
struct node *nextPtr;
}*LinkList, Lnode;

static void CreateList(LinkList *headPtr, LinkList *tailPtr, char ch);
static void Decompose(LinkList *headPtrA, LinkList *headPtrB, LinkList *tailPtrB);
static void VisitList(LinkList headPtr);
static void DestroyList(LinkList *headPtr, LinkList *tailPtr);

int main(void)
{
LinkList headPtrA = NULL, tailPtrA = NULL, headPtrB = NULL, tailPtrB = NULL;
char ch;

while (1)
{
printf("Enter ch('@'-quit): ");
scanf(" %c", &ch);
if (ch == '@')
{
break;
}
else
{
CreateList(&headPtrA, &tailPtrA, ch);
}
}

VisitList(headPtrA); /* 打印分解前的链表 */

if (headPtrA != NULL) /* 链表不空的情况对其进行分解 */
{
Decompose(&headPtrA, &headPtrB, &tailPtrB); /* 对链表进行分解*/
}
else
{
printf("headPtrA is empty.\n");
}

VisitList(headPtrA); /* 打印分解后的链表 */
VisitList(headPtrB);

DestroyList(&headPtrA, &tailPtrA); /* 销毁链表 */
DestroyList(&headPtrB, &tailPtrB);

return 0;
}

static void CreateList(LinkList *headPtr, LinkList *tailPtr, char ch)
{
LinkList newPtr;

if ((newPtr = (LinkList)malloc(sizeof(Lnode))) == NULL)
{
exit(1);
}

newPtr -> data = ch;
newPtr -> nextPtr = NULL;

if (*headPtr == NULL)
{
newPtr -> nextPtr = *headPtr;
*headPtr = newPtr;
}
else
{
(*tailPtr) -> nextPtr = newPtr;
}
*tailPtr = newPtr;
}

static void Decompose(LinkList *headPtrA, LinkList *headPtrB, LinkList *tailPtrB)
{
int count = 0;
LinkList cA, pA;
char ch;

cA = NULL;
for (pA = *headPtrA; pA != NULL; cA = pA,pA = pA -> nextPtr)
{
ch = pA -> data;
count++;
if (count % 2 == 0)
{
CreateList(headPtrB, tailPtrB, ch);
cA -> nextPtr = pA -> nextPtr;
}
}
}

static void VisitList(LinkList headPtr)
{
while (headPtr != NULL)
{
printf("%c -> ", headPtr -> data);
headPtr = headPtr -> nextPtr;
}
printf("NULL\n");
}

static void DestroyList(LinkList *headPtr, LinkList *tailPtr)
{
LinkList tempPtr;

while (*headPtr != NULL)
{
tempPtr = *headPtr;
*headPtr = (*headPtr) -> nextPtr;
free(tempPtr);
}
*headPtr = NULL;
*tailPtr = NULL;
}

我想了个算法 不知道对不对 如果错了请改正
就是每次节点往后走两个
可以用for循环
for(head1=head->next;head1!=NULL;head1->next->next)
{
head1->next=head1->next->next;
}
for(head2=head->next->next;head2!=NULL;head2->next->next)
{
head2->next=head2->next->next;
}