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;
}
}