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
     *
     *
     * @param head ListNode类
     * @return ListNode类
     */
    ListNode* oddEvenList(ListNode* head) {
        // write code here
        if (head == nullptr || head->next == nullptr || head->next->next == nullptr) {
            return head;
        }

        ListNode* odd_head = head;
        ListNode* even_head = head->next;
        auto p = odd_head;
        auto q = even_head;
        while (p->next != nullptr && q->next != nullptr) {
            if (q->next != nullptr) {
                p->next = q->next;
                p = p->next;
                if (p->next != nullptr ) {
                    if (p->next != nullptr) {
                        q->next = p->next;
                        q = q->next;
                    } else {
                        break;
                    }
                }
            } else {
                break;
            }
        }

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

        p->next = even_head;
        return odd_head;
    }
};