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 。

提示：

• 链表中节点的数目在范围 [0, $5\ast 10^4$] 内
• $-10^5$ <= Node.val <= $10^5$

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？

建议用vscode时间测试，分别转成数组排序和只在链表基础上操作

题解1 STL - multiset

/**
 * 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) {
        if(!head || !head->next) return head;
        ListNode* subhead = head;
        multiset<int> kkset;
        while(subhead){
            kkset.insert(subhead->val);
            subhead = subhead->next;
        }
        subhead = head;
		// 直接改值
        for(auto& i : kkset){
            subhead->val = i;
            subhead = subhead->next;
        }
        subhead = nullptr;
        return head;

    }
};

题解2 归并【自顶向下】

class Solution {
public:
	// 递归
    ListNode* mergesort(ListNode* head, ListNode* tail){
        if(! head || head->next == tail){
            // 不含tail
            if(head) head->next = nullptr;
            return head;
        }
        // 找到head和tail之间的中点(slow)
        ListNode *fast, *slow;
        fast = slow = head;
        while(fast != tail){
            slow = slow->next;
            fast = fast->next;
            if(fast != tail)
                fast = fast->next;
        }
        // ListNode* midnode = slow;
        // recursion
        // return  merge(mergesort(head, midnode),  mergesort(midnode, tail));
		return  merge(mergesort(head, slow),  mergesort(slow, tail));
    }
    ListNode* sortList(ListNode* head) {
        return mergesort(head, nullptr);
    }
    
    // 21.合并两个有序链表
    ListNode* merge(ListNode*l1, ListNode* l2){
        ListNode* dummynode = new ListNode(0);
        ListNode *p = dummynode;
        dummynode->next = p;
        
        while(l1 || l2){

            if(l1){
                if(l2 && l1->val <= l2->val){
                    p->next = l1;
                    p = p->next;
                    l1 = l1->next;
                    continue;
                }
                if(!l2){
                    p->next = l1;
                    break;
                }
            }

            if(l2){
                if(l1 && l2->val < l1->val){
                    p->next = l2;
                    p = p->next;
                    l2 = l2->next;
                    continue;
                }
                if(!l1){
                    p->next = l2;
                    break;
                }
            }
        }
        return dummynode->next;
    }
};

题解3 归并【自底向上】

自底向上：子串长度 l 从1开始，合并后的串长度*2，1+1 -> 2+2 -> 4+4 ->…

// 有时间自己再写一遍
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;
    }
};