willsun888
5/14/2013 - 4:53 AM

反转一个单链表,分别以迭代和递归的形式来实现

反转一个单链表,分别以迭代和递归的形式来实现

// 返回新链表头节点
LinkNode *reverse_link(LinkNode *head)
{
  if(head == NULL)
		return NULL;
	LinkNode *prev , *curr , *reverse_head , *temp;
	prev = NULL , curr = head;
	while(curr->next)
	{
		temp = curr->next;
		curr->next = prev;
		prev = curr;
		curr = temp;
	}
	curr->next = prev;
	reverse_head = curr;
	return reverse_head;
}

LinkNode *reverse_link_recursive(LinkNode *head)
{
	if(head == NULL)
		return NULL;
	LinkNode *curr , *reverse_head , *temp;
	if(head->next == NULL)    // 链表中只有一个节点,逆转后的头指针不变
		return head;
	else
	{
		curr = head;
		temp = head->next;    // temp为(a2,...an)的头指针
		reverse_head = reverse_link_recursive(temp);   // 逆转链表(a2,...an),并返回逆转后的头指针
		temp->next = curr;    // 将a1链接在a2之后
		curr->next = NULL;
	}
	return reverse_head;      // (a2,...an)逆转链表的头指针即为(a1,a2,...an)逆转链表的头指针
}