题解 | #链表的奇偶重排#

/**
 * 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
     *
     * 该题的关键是用笔画出结点之间的关系
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* oddEvenList(ListNode* head) {
        if (head == nullptr || head->next == nullptr) return head;
        auto p = head;
        auto head1 = head;
        auto q = head->next;
        auto head2 = q;

        while (q->next != nullptr && q->next->next != nullptr) {
          p->next = q->next;
          p = q->next;
          q->next = p->next;
          q = q->next;
        }

        if (q->next == nullptr) {
          p->next = nullptr;
          q->next = nullptr;
        } else {
          p->next = q->next;
          p = p->next;
          q->next = nullptr;
        }

        p->next = head2;
        return head1;
    }
};