题解 | #重排链表#快慢指针分为前后半段，然后翻转后半段

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
  public:
    void reorderList(ListNode* head) {
        if (!head || !head->next) return;
        ListNode* slow = head, *qulick = head->next;
        while (qulick && qulick->next) {
            slow = slow->next;
            qulick = qulick->next->next;
        }
        ListNode* left = head, *right = slow->next;
        slow->next = nullptr;
        right = reverse(right);
        ListNode* p = left, *q = right, *dummy = new ListNode(-1), *tail = dummy;
        while (p && q) {
            tail->next = p;
            tail = p;
            p = p->next;
            tail->next = q;
            tail = q;
            q = q->next;
        }
        if (p) tail->next = p;
    }

    ListNode* reverse(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode* newHead = reverse(head->next);
        head->next->next = head;
        head->next = nullptr;
        return newHead;
    }
};