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||!head->next) {
            return head;
        }
        
        ListNode* head1, *head2, *tail1, *tail2;
        ListNode* p = head;
        int n=1;
        if (p) {
            head1 = p;
            tail1 = p;
            p = p->next;
            n++;
        }
        if (p) {
            head2 = p;
            tail2 = p;
            p = p->next;  
            n++;          
        }

        while (p) {
            if (n % 2) {
                tail1->next = p;
                tail1 = tail1->next;
            } else {
                tail2->next = p;
                tail2 = tail1->next;
            }
            n++;
            p = p->next;
        }
        tail1->next = head2;
        tail2->next = nullptr;
        return head1;

    }
};