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 。

题解1 线性表

/*** 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) return head;ListNode* dummynode = new ListNode(-1);ListNode* p = dummynode;p->next = head;vector<int> arr;while(head){arr.push_back(head->val);head = head->next;}sort(arr.begin(), arr.end());int i = 0;while(p->next){p->next->val = arr[i++];p = p->next;}return dummynode->next;}
};

题解2 自顶向下归并排序

class Solution {
public:ListNode* sortList(ListNode* head) {return sortList(head, nullptr);}ListNode* sortList(ListNode* head, ListNode* tail) {if (head == nullptr) {return head;}if (head->next == tail) {head->next = nullptr;return head;}// 快慢指针找midListNode* slow = head, *fast = head;while (fast != tail) {slow = slow->next;fast = fast->next;if (fast != tail) {fast = fast->next;}}ListNode* mid = slow;return merge(sortList(head, mid), sortList(mid, tail));}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;}
};

题解3 自底向上归并排序

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);// 从1 -> 2 -> 4 -> ··· -> for (int subLength = 1; subLength < length; subLength <<= 1) {ListNode* prev = dummyHead, *curr = dummyHead->next;while (curr != nullptr) {ListNode* head1 = curr;// 断成2个sublength的表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;}// 合并两个表ListNode* merged = merge(head1, head2);prev->next = merged;while (prev->next != nullptr) {prev = prev->next; // 到下一个prev（merged尾）}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;}
};