因为只需要维护3个指针，所以空间复杂度为 $O(1)$ 。

class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if (!head) return head;
        ListNode* evenHead = head->next;        //奇头是head，偶头就是head->next
        ListNode* odd = head, *even = evenHead; //分别定义奇偶尾指针

        while (even && even->next)
        {
            odd = odd->next = even->next;       //等于 odd->next = even->next; odd = odd->next;
            even = even->next = odd->next;
        }

        odd->next = evenHead;

        return head;
    }
};