①：用 vector 存下整個連結串列，然後按題目要求鏈起來即可。時間空間 O(n)

②：將連結串列從中間位置分割為兩個連結串列，並將後半部分連結串列進行反轉，然後再鏈起來即可。時間 O(n)，空間 O(1)

class Solution {
public:
    void reorderList(ListNode* head) {
        ListNode *tail = reverseHalfList(head);
        ListNode *ptr = head;
        while(tail) {      // tail 為 nullptr, 整個連結就完成了.
 

            ListNode *ptrNext = ptr -> next;
            ListNode *tailLast = tail -> next;
            ptr -> next = tail;
            tail -> next = ptrNext;
            ptr = ptrNext;
            tail = tailLast;
        }
    }

private:
    ListNode* reverseHalfList(ListNode *head) {
        ListNode 
 
*slow = head, *quick = head;
        // 尋找連結串列中間節點
        // 連結串列結點為單數, slow 停止中間節點. 為偶數 slow 停止在中間第一個結點
        while(quick && quick -> next && quick -> next -> next) {
            slow = slow -> next;
            quick = quick -> next -> next;
        }

        ListNode 
 
*slowNext = slow -> next;
        slow -> next = nullptr;
        return reverseList(slowNext);
    }

    ListNode* reverseList(ListNode *head) {
        ListNode *pre = nullptr;
        while(head) {
            ListNode *nxt = head -> next;
            head -> next = pre;
            pre = head;
            head = nxt;
        }
        return pre;
    }
};