挑战面试编程：链表逆转的多种实现

#include <stdio.h>
#include <stdlib.h>
//链表节点
typedef struct node
{
int data;
struct node *next;
}Node;
//逆转的while实现
void rev(Node **list)   //链表逆转
{
if (*list == NULL || (*list)->next == NULL)  //空或者仅仅有一个节点
{
return;
}
else
{
Node *pPre, *pCur, *pNext;

pPre = *list;
pCur = (*list)->next;

while (pCur)
{
pNext = pCur->next;

pCur->next = pPre;

pPre = pCur;

pCur = pNext;

}

(*list)->next = NULL;
*list = pPre;
}
}
//逆转的递归实现
void rev_r(Node **list, Node *pCur, Node *pNext)
{
if (pCur == NULL || pNext == NULL)
{
(*list)->next = NULL;

*list = pCur;

return;
}

rev_r(list, pNext, pNext->next);

pNext->next = pCur;
}
//逆转的goto实现
void rev_goto(Node **list)
{
if (*list == NULL || (*list)->next == NULL)
return;
else
{
Node *pPre = *list;
Node *pCur = (*list)->next;
Node *pNext = NULL;

stage:
pNext = pCur->next;

pCur->next = pPre;

pPre = pCur;

pCur = pNext;

if (pCur)
goto stage;

(*list)->next = NULL;
*list = pPre;
}
}
void print(Node *list)
{
if (list == NULL)
return;
printf("%4d", list->data);
print(list->next);
}
void main()
{
Node p1 = { 1, NULL };
Node p2 = { 2, &p1 };
Node p3 = { 3, &p2 };
Node p4 = { 4, &p3 };
Node p5 = { 5, &p4 };
Node p6 = { 6, &p5 };
Node *list = &p6;
printf("原链表\n");
print(list);
printf("\n");
printf("逆转\n");

//rev(&list);   //while
//rev_r(&list, list, list->next);   //递归
rev_goto(&list);   //goto
print(list);
printf("\n");
system("pause");
}