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 。

示例 1：

输入：head = [4,2,1,3]
输出：[1,2,3,4]

示例 2：

输入：head = [-1,5,3,4,0]
输出：[-1,0,3,4,5]

示例 3：

输入：head = []
输出：[]

提示：

• 链表中节点的数目在范围 [0, 5 * 10⁴] 内
• -10⁵ <= Node.val <= 10⁵

参考答案

class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (head == nullptr) {
            return head;
        }
        int length = 0;
        ListNode* node = head;
        while (node != nullptr) {
            length++;
            node = node->next;
        }
        ListNode* dummyHead = new ListNode(0, head);
        for (int subLength = 1; subLength < length; subLength <<= 1) {
            ListNode* prev = dummyHead, *curr = dummyHead->next;
            while (curr != nullptr) {
                ListNode* head1 = curr;
                for (int i = 1; i < subLength && curr->next != nullptr; i++) {
                    curr = curr->next;
                }
                ListNode* head2 = curr->next;
                curr->next = nullptr;
                curr = head2;
                for (int i = 1; i < subLength && curr != nullptr && curr->next != nullptr; i++) {
                    curr = curr->next;
                }
                ListNode* next = nullptr;
                if (curr != nullptr) {
                    next = curr->next;
                    curr->next = nullptr;
                }
                ListNode* merged = merge(head1, head2);
                prev->next = merged;
                while (prev->next != nullptr) {
                    prev = prev->next;
                }
                curr = next;
            }
        }
        return dummyHead->next;
    }

    ListNode* merge(ListNode* head1, ListNode* head2) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* temp = dummyHead, *temp1 = head1, *temp2 = head2;
        while (temp1 != nullptr && temp2 != nullptr) {
            if (temp1->val <= temp2->val) {
                temp->next = temp1;
                temp1 = temp1->next;
            } else {
                temp->next = temp2;
                temp2 = temp2->next;
            }
            temp = temp->next;
        }
        if (temp1 != nullptr) {
            temp->next = temp1;
        } else if (temp2 != nullptr) {
            temp->next = temp2;
        }
        return dummyHead->next;
    }
};