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 == NULL) return NULL;
        if (head->next == NULL)return head;
        if (head->next->next == NULL)return head;

        ListNode* odd = head, * even = head->next;
        ListNode* lefthead = new ListNode(0), * righthead = new ListNode(0);
        ListNode* l = lefthead, * r = righthead;
        ListNode* ltail, * rtail;
        while (odd || even) {
            ltail = odd;
            rtail = even;
            if (odd->next) { //这里不用判断odd是否存在，原因是只要能进循环odd就必然存在，而even不一定
                odd = odd->next->next;
            } else {
                odd = odd->next;
            }
            if (even) {
                if (even->next) {
                    even = even->next->next;
                } else {
                    even = even->next;
                }
            }
            if (ltail) {
                ltail->next = NULL;
                l->next = ltail;
                l = l->next;
            }
            if (rtail) {
                rtail->next = NULL;
                r->next = rtail;
                r = r->next;
            }
        }
        ltail->next = righthead->next;
        return lefthead->next;
    }
};

采用奇偶双指针的思路，沿着链表以步长为2前进，每前进一次就使上一步的结点断裂出去，新建两个头指针用于接收断裂下来的结点，最后合并两个链表。空间复杂度为常数，时间复杂度为On。