思路

首先将链表一份为二，首先对链表进行计数，cnt。如果cnt小于等于二说明不用更改直接返回。
否则找到将链表一分为二，循环(cnt + 1) / 2-1次，得到前一个链表的最后一个节点mid。
获取到下一个链表的第一个结点later，再把mid的next变为nullptr，对later进行反转。
反转前将cnt赋值为0，计算后一链表的长度最后决定应该执行几次插入操作。

C++ 代码

class Solution {
public:
    void reorderList(ListNode* head) {
        int cnt = 0;
        for (auto i = head; i; i = i->next) cnt ++;
        if (cnt <= 2) return;

        auto mid = head;
        for (int i = 1; i < (cnt + 1) / 2;i ++) mid = mid->next;

        cnt = 0;
        auto later = mid->next;
        mid->next = nullptr;
        ListNode* pre = nullptr;
        while (later){
            auto ne = later->next;
            later->next = pre;
            pre = later, later = ne;
            cnt ++;
        }

        for (int i = 0; i < cnt; i ++){
            later = pre->next;
            pre->next = head->next;
            head->next = pre;
            head = pre->next;
            pre = later;
        }

    }
};