题目表述：将给定的单链表L： L_0→L_1→…→L_{n-1}→L_ nL0​→L1​→…→Ln−1​→Ln​
重新排序为：L_0→L_n →L_1→L_{n-1}→L_2→L_{n-2}→…L0​→Ln​→L1​→Ln−1​→L2​→Ln−2​→…
要求使用原地算法，不能只改变节点内部的值，需要对实际的节点进行交换。
 

示例：

输入：{1,2,3,4}   返回值：{1,4,2,3}

给定head链表1->2->3->4, 重新排列为 1->4->2->3,会取head链表里面的值打印输出

解题思路：首先利用快慢指针，找出链表中间端，将链表一分为二。将后半段链表翻转，再将两个列表依次按序连接即可。

void reorderList(ListNode *head) {  //重排链表
        if(!head) return ;
//找到中间节点
        ListNode *fast,*slow;
        fast = head,slow = head;
        while (fast->next != NULL && fast->next->next != NULL)
        {
            fast = fast->next->next;
            slow = slow->next;
        }
//拆分链表
        ListNode *first, *second ; 
        first = head, second = slow->next;
        slow->next = NULL;
//反转链表
        ListNode *current,*temp,*newh;
        newh = NULL;
        current = second;
        while (current != NULL)
        {
            temp = newh;
            newh = current;
            current = current->next;
            newh->next = temp;
        }
        second = newh;
//合并两个链表
        ListNode *first_1,*second_1;
        while((first!=NULL)&&(second!=NULL))
        {
            first_1 = first->next;
            second_1 = second->next;
            first->next = second;
            first = first_1;
            second->next = first;
            second = second_1;
        }
    }