题解 | #单链表的排序#记录

/**
 * 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 the head node
     * @return ListNode类
     */
    ListNode* sortInList(ListNode* head) {
        // write code here
        vector<ListNode*> lists;
        ListNode *p;
        while(head){
            p = head->next;
            head->next = nullptr;
            lists.push_back(head);
            head = p;
        }

        return div(lists, 0, lists.size()-1);
    }
	//归并排序
    ListNode* div(vector<ListNode*>& lists, int l, int r) {
        if (l > r) return nullptr;
        if (l == r) return lists[l];

        int m = (l + r) / 2;
        return Merge(div(lists, l, m), div(lists, m + 1, r));
    }

    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        // write code here

        ListNode* head;
        ListNode* p2, *p1, *s;
        if (pHead1 && pHead2) {
            if (pHead1->val > pHead2->val) {
                head = pHead1;
                p2 = pHead2;
            } else {
                head = pHead2;
                p2 = pHead1;
            }
        } else if (pHead1) {
            return pHead1;
        } else return pHead2;
        p1 = head;

        //双指针，p1,p2，比大小前插，小则后移
        head = p2;
        p2 = p2->next;
        head->next = p1;
        s = head;
        while (p2) {
            //
            if (p2->val <= p1->val) {
                s->next = p2;
                p2 = p2->next;
                s = s->next;
                s->next = p1;
            } else {
                p1 = p1->next;
                s = s->next;
                if (!p1) {
                    s->next = p2;
                    return head;
                }
            }
        }

        return head;
    }
};

时间复杂度和空间复杂度应该都符合题意。

准备尝试快速排序，防丢失