[数据结构与算法]链表的反转(不使用栈)

将链表进行反向输出:

#include <iostream>

typedef struct listnode {
	int data;
	struct listnode* next;
}listnode,*list;

void reserve(listnode *node)
{
	//头部的后继节点置为空
	//在转换后继节点的时候，需要提前将后继节点取出

	if (node != nullptr)
	{
		listnode *head_node = new listnode;
		head_node->data = 0;
		head_node->next = node;
		listnode *tmp_node = nullptr;
		tmp_node = head_node->next;
		while (tmp_node)
		{
			listnode *next_node = tmp_node;//当前节点的后继节点
			tmp_node = tmp_node->next;	//原当前节点的后后继节点
			next_node->next = head_node; //后继节点指针反转
			head_node = next_node;//node升级为后继节点
		}
		while (head_node)
		{
			printf("%d", head_node->data);
			head_node = head_node->next;
			if (head_node->data == 0)
			{
				head_node = nullptr;
			}
		}
	}
}
int main()
{
	listnode *next_node1 = new listnode;
	listnode *next_node2 = new listnode;
	listnode *next_node3 = new listnode;
	listnode *next_node4 = new listnode;
	listnode *next_node5 = new listnode;

	next_node1->data = 1;
	next_node1->next = next_node2;
	next_node2->data = 2;
	next_node2->next = next_node3;
	next_node3->data = 3;
	next_node3->next = next_node4;
	next_node4->data = 4;
	next_node4->next = next_node5;
	next_node5->data = 5;
	next_node5->next = nullptr;
	reserve(next_node1);
}