用两种递归思路与循环实现单链表的反转

typedef struct ListNode{
	int data;
	struct ListNode *next;
}ListNode;

//递归一
ListNode *ReverseList (ListNode *pHead, ListNode *nHead = NULL)
{
	//每次取下第一个节点头插法创建新链表
	//nHead为反转后链表的头节点
	if(pHead == NULL)
	  return NULL;
	ListNode *pNext = pHead -> next;
	pHead -> next = nHead;
	nHead = pHead;

	if (pNext == NULL)
	  return nHead;
	else
	  return ReverseList (pNext, nHead);
}

//递归二
ListNode *ReverseList (ListNode *pHead)
{
	//递归到链表的最后一个将其指针反转
	if (pHead == NULL || pHead -> next == NULL)
	  return pHead;
	else
	{
		ListNode *nHead = ReverseList (pHead -> next);
		//将最后两个节点反转
		pHead ->next ->next = pHead;
		pHead ->next = NULL;

		return nHead; //这是反转后链表的头节点
	}
}

//循环
ListNode *ReverseList (ListNode *pHead)
{
	//每次取下第一个节点头插法构造新链表
	ListNode *nHead = NULL;
	ListNode *pNode = pHead; //当前取下的节点
	ListNode *prev = NULL;

	while (pNode != NULL)
	{
		ListNode *pNext = pNode -> next;
		if (pNext == NULL)
		  nHead = pNode;   //到达链表的最后一个节点
		pNode -> next = prev;
		prev = pNode;
		pNode = pNext;
	}
	
	return nHead;
}

用两种递归思路与循环实现单链表的反转,布布扣,bubuko.com

用两种递归思路与循环实现单链表的反转

原文：http://blog.csdn.net/willib/article/details/38386547