148. 排序链表 链表归并排序

给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。

进阶：

你可以在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sort-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        ListNode* dummy = new ListNode;
        dummy->next = head;
        
        auto p = head;
        int len = 0;
        while (p) {
            len++;
            p = p->next;
        }

        for (int i = 1; i < len; i <<= 1) {
            auto cur = dummy->next;
            auto tail = dummy;

            while (cur) {
                auto l = cur;
                auto r = cut(l, i);
                cur = cut(r, i);
                tail->next = merge(l, r);
                while (tail->next) {
                    tail = tail->next;
                }
            }
        }

        return dummy->next;
    }

    ListNode* cut(ListNode* head, int n) {
        auto p = head;
        while (--n && p) {
            p = p->next;
        }

        if (!p) {
            return nullptr;
        }

        auto next = p->next;
        p->next = nullptr;
        return next;
    }

    ListNode* merge(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode;
        auto p = dummy;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                p->next = l1;
                p = l1;
                l1 = l1->next;
            }
            else {
                p->next = l2;
                p = l2;
                l2 = l2->next;
            }
        }
        p->next = l1 ? l1 : l2;
        return dummy->next;
    }


};