http://bbs.chinaunix.net/forum/23/20041101/436188.html
GAORJ 的问题如下：
有一单链表n1->n2->...n(x-1)->nx,请问如何在不使用辅助节点的情况下实现该链表的反转呢? 反转后变成nx->n(x-1)->...n2->n1 .
（随机测试，最大最至20000，都能有正确的结果）
代码如下：
#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;
}
