题意：

　　给你一个单链表 a1 a2 a3 a4 a5....an　　让你变成 a1 an a2 a_n-1 a3 a_n-2 ....

这里牵涉到，单链表的倒置和两个单链表的合并。

class Solution {public:    void reorderList(ListNode *head) {        if (!head || !head->next)return;    //空表和只有一个元素的链表不用操作        //快慢指针找中间位置        ListNode *fast = head;        ListNode *slow = head;        while (fast&&fast->next)        {            slow = slow->next;    fast = fast->next->next;        }        //将后半段的链表倒置        ListNode *after = slow->next;    slow->next = NULL;        ListNode *pre = NULL;        while (after){            ListNode *p = after->next;            after->next = pre;            pre = after;    after = p;        }        //合并两个链表        ListNode *first = head;        after = pre;        while (first&&after){            ListNode *ftemp = first->next;            ListNode *aftemp = after->next;            first->next = after;    first = ftemp;            after->next = first;    after = aftemp;        }    }};