题解 | #链表的奇偶重排#

class Solution {
  public:
    ListNode* oddEvenList(ListNode* head) {
        if (head == nullptr || head->next == nullptr || head->next->next == nullptr) {
            return head;
        }
        ListNode* oddListHead = new ListNode(head->val);
        ListNode* oddList = oddListHead;
        ListNode* evenListHead = new ListNode(head->next->val);
        ListNode* evenList = evenListHead;
        auto p = head;
        while (p->next != nullptr && p->next->next != nullptr) {
            p = p->next->next;
            auto q = new ListNode(p->val);
            oddList->next = q;
            oddList = q;
        }
        p = head->next;
        while (p->next != nullptr && p->next->next != nullptr) {
            p = p->next->next;
            auto q = new ListNode(p->val);
            evenList->next = q;
            evenList = q;
        }
        oddList->next = evenListHead;
        return oddListHead;
    }
};