有一單鏈表n1->n2->...n(x-1)->nx,請問如何在不使用輔助節點的情況下實現該鏈表的反轉呢? 反轉后變成nx->n(x-1)->...n2->n1 .

<!--StartFragment --> （隨機測試，最大最至20000，都能有正確的結果） 

代碼如下：

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

#define SWAP(a,b) { \
a=a^0xFFFFFFFF; \
b=b^a; \
a=a^b; \
b=b^0xFFFFFFFF; \
b=b^a; \
}
#define SWAP_POINT(pA,pB) {\
int h,t; \
pLinkList ph,pt; \
h=(int)pA; \
t=(int)pB; \
ph = (pLinkList)h; \
pt = (pLinkList)t; \
SWAP(h,t); \
pA = (pLinkList)h; \
pB = (pLinkList)t; \
}



typedef struct tagLinkList
{
int data;
struct tagLinkList *next;
} LinkList , *pLinkList ;




int main(int argc, char* argv[])
{
int count,i;
pLinkList Head,tmpNode,Tail;

Head=NULL;

// Construct LinkList.
printf("Enter count:"); scanf("%d",&count);
if(count <= 1) 
{
printf("Count error,must large than 1.\n");
exit(1);
}
for(i=1;i<=count;++i)
{
tmpNode=(LinkList *)malloc(sizeof(LinkList));
tmpNode->data=i;
printf("No.=%d\t Address:%X , Data:%d\n",i,tmpNode,tmpNode->data);
if(!Head)
Tail=Head=tmpNode;

else{
Tail->next=tmpNode;
Tail=tmpNode;
}
}
Tail->next=NULL;

// Reverse LinkList.
if(2 == count) {
//N=2
SWAP_POINT(Head,Tail->next);
SWAP_POINT(Head,Tail);
SWAP_POINT(Head->next->next,Tail);
}else if(3 == count) {
//N=3
SWAP_POINT(Head->next,Tail->next);
SWAP_POINT(Head,Tail);
SWAP_POINT(Head->next->next,Tail);
}else{
//N>=4
SWAP_POINT(Head->next,Tail->next);
SWAP_POINT(Head,Tail);
SWAP_POINT(Head->next->next,Tail);
SWAP_POINT(Head->next,Tail->next);
SWAP_POINT(Head->next->next,Tail);
for(i=6;i<=count;++i)
{
SWAP_POINT(Head->next,Tail);
SWAP_POINT(Head->next->next,Tail);
}
}

//Output reversed LinkList.
tmpNode=Head;
printf("After reverse.\n");
for(i=1;i<=count;++i)
{
printf("No.=%d\t Address:%X , Data:%d\n",i,tmpNode,tmpNode->data);
tmpNode = tmpNode->next;
}

//Release LinkList
printf("Release LinkList.\n");
for(i=1;i<=count;++i)
{
tmpNode=Head;
Head = Head->next;
free(tmpNode);
}



return 0;
}