题目描述

题目的进阶问题要求达到 O(nlogn) 的时间复杂度和 O(1) 的空间复杂度，时间复杂度是 O(nlogn) 的排序算法包括归并排序、堆排序和快速排序（快速排序的最差时间复杂度是 O(n2），其中最适合链表的排序算法是归并排序

法一 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;}ListNode *fast=head, *slow = 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 *l1, ListNode *l2){ListNode *dummy = new ListNode(-1);ListNode *cur=dummy, *s1=l1, *s2=l2;while(s1 && s2){if(s1->val < s2->val){cur->next = s1;s1 = s1->next;} else {cur->next = s2;s2 = s2->next;}cur = cur->next;}cur->next = s1 ? s1 : s2;return dummy->next; }
};

法二）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);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;}
};