1. 题目

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。

示例 1:输入: 4->2->1->3
输出: 1->2->3->4
示例 2:输入: -1->5->3->4->0
输出: -1->0->3->4->5

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sort-list
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2. 解题

2.1 归并排序

参考单链表归并排序

class Solution {
public:ListNode* sortList(ListNode* head) {if(head == NULL || head->next == NULL)return head;ListNode *fast = head->next, *slow = head, *rightHead;//fast初始化先走一步，偶数个链表时，防止一边0个，一边2个节点while(fast && fast->next){fast = fast->next->next;slow = slow->next;}rightHead = slow->next;slow->next = NULL;//先断开，再操作左右！！！ListNode *right = sortList(rightHead);ListNode *left = sortList(head);return merge(left, right);}ListNode* merge(ListNode *left, ListNode *right) {ListNode *emptyHead = new ListNode(0);//虚拟哨兵头ListNode *cur = emptyHead;while(left && right){if(left->val < right->val){cur->next = left;left = left->next;}else{cur->next = right;right = right->next;}cur = cur->next;}cur->next = (left == NULL ? right : left);cur = emptyHead->next;delete emptyHead;return cur;}
};

2.2 快速排序

class Solution {
public:ListNode* sortList(ListNode *head) {quicksort(head, NULL);return head;}void quicksort(ListNode *head, ListNode *tail) {if(head == tail || head->next == NULL)return;ListNode *mid = partition(head,tail);quicksort(head, mid);quicksort(mid->next, tail);}ListNode* partition(ListNode *head, ListNode *tail){int pivot = head->val;ListNode *left = head, *cur = head->next;while(cur != NULL && cur != tail){if(cur->val < pivot){left = left->next;swap(cur->val, left->val);}cur = cur->next;}swap(left->val, head->val);return left;}
};