面试中常见链表问题8：重排链表

    给定一个单链表：L0->L1->...->Ln-1->Ln，重新排序以后返回L0->Ln->L1->Ln-1->L2->Ln-2...。
    解析：(1)把单链表分割为前后两部分。(2)把后一部分链表反转。(3)把两部分链表交叉插入即可。

ListNode * reverseList(ListNode *head)
{
	if (head == NULL || head->next == NULL)
		return head;
	ListNode *pHead = new ListNode(INT_MAX);
	pHead->next = head;
	ListNode *p = head->next;
	head->next = NULL;
	while (p)
	{
		ListNode *tmp = p;
		p = p->next;
		tmp->next = pHead->next;
		pHead->next = tmp;
	}
	return pHead->next;
}

void reorderList(ListNode* &head) 
{
	if (head == NULL || head->next == NULL)
		return;
	ListNode *slow = head;
	ListNode *fast = head;
	while (fast != NULL && fast->next != NULL)
	{
		fast = fast->next->next;
		slow = slow->next;
	}
	ListNode *pHead1 = head;
	ListNode *pHead2 = slow->next;
	slow->next = NULL;
	pHead2 = reverseList(pHead2);
	while (pHead2 != NULL)
	{
		ListNode *tmp2 = pHead1->next;
		ListNode *tmp = pHead2;
		pHead2 = pHead2->next;
		tmp->next = pHead1->next;
		pHead1->next = tmp;
		pHead1 = tmp2;
	}
}