题解 | #链表的奇偶重排#

思路：
1.将链表拆分为奇节点链表和偶节点链表；
2.找到奇节点链表结尾，连接上偶节点链表并返回


/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @return ListNode类
     */
    ListNode* oddEvenList(ListNode* head) {
        // write code here
        if (!head || !head->next) return head;
        ListNode* newHead = new ListNode(-1), *p = head, *q, *temp;
        q = newHead;
        while (p && p->next) {
            temp = p->next;
            p->next = p->next->next;
            p = p->next;
            q->next = temp;
            q = q->next;
        }
        // if(!p) p = newHead->next;
        // else p->next = newHead->next;
        q->next = nullptr;
        p = head;
        while (p->next) {
            p = p->next;
        }
        p->next = newHead->next;
        delete newHead;
        return head;
    }
};