单链表的基本处理，涉及逆序，计算长度等操作。
题意要求：
1->2->3->4 => 1->4->2->3 间隔操作。
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

class Solution {
public:
    void reorderList(ListNode *head) {
        int len = getLength(head);
        if(len <= 2)return ;
        int pos = len%2==0?len/2:len/2+1;
        ListNode * headend = getListNode(head,pos-1);
        ListNode * head2 = getListNode(head,pos);
        headend -> next = NULL;
        head2 = reverseListNode(head2);
        ListNode * headc = head;
        while(head2 != NULL) {
            ListNode * headnext = head->next;
            ListNode * head2next = head2->next;
            head->next = head2;
            head2->next = headnext;
            head = headnext;
            head2 = head2next;
        }
        head = headc;
    }
private:
    ListNode * reverseListNode(ListNode * head) {
        ListNode * before = head;
        ListNode * now = head->next;
        while(now != NULL) {
            ListNode * next = now->next;
            now->next = before;
            before = now;
            now = next;
        }
        head->next = NULL;
        return before;
    }
    int getLength(ListNode *head) {
        int len = 0;
        while(head != NULL) {
            len++;
            head = head->next;
        }
        return len;
    }
    ListNode *getListNode(ListNode *head,int pos) {
        int i = 0;
        while(i < pos) {
            head = head->next;
            i++;
        }
        return head;
    }
};