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)  //空或者只有一个节点
	{
		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");
}