[数据结构与算法]反转链表——迭代法

1->2->3->4->5反转为：5->4->3->2->1

#include <stdio.h>
#include<stdlib.h>
struct node
{
	int num;
	struct node *next;
};
void PrintLink(struct node *p)
{
	struct node *head=p;
	while(head!=NULL)
	{
		printf("%d ",head->num);
		head=head->next;
	}
	putchar('\n');
}
int totalNode(struct node *p)
{
	struct node *head=p;
	int num=0;
	while(head!=NULL)
	{
		num++;
		head=head->next;
	}
	return num;

}
struct node *fanzhuan(struct node *phead)//反转链表操作——迭代法
{
	struct node *head=phead;

	struct node *b;
	struct node *a;
	if(head==NULL)
	{
		return head;
	}
	else
	{
		a=phead;
		head=NULL;
		while(a!=NULL)
		{
			b=a->next;
			a->next=head;
			head=a;	

			a=b;
		}
	}
	return head;


}
int main()
{
	struct node t5={5,NULL};
	struct node t4={4,NULL};
	struct node t3={3,NULL};
	struct node t2={2,NULL};
	struct node t1={1,NULL};

	t1.next=&t2;
	t2.next=&t3;
	t3.next=&t4;
	t4.next=&t5;

	PrintLink(&t1);
	printf("Total Node: %d\n",totalNode(&t1));
	struct node *pp=fanzhuan(&t1);
	PrintLink(pp);
	return 0;
}

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