题目链接：https://leetcode.cn/problems/sort-list/?envType=study-plan-v2&envId=top-100-liked

第一种，插入排序，会超时

class Solution {public ListNode sortList(ListNode head) {//插入排序，用较为简单的方式解决ListNode dummyNode = new ListNode(0,null);ListNode p = head;while(p!=null){ListNode temp = p;p=p.next;//将temp插入到链表当中ListNode q = dummyNode;int flag = 0;while(q.next!=null){if(q.next.val>=temp.val){ListNode t = q.next;q.next = temp;temp.next = t;flag= 1;break;}q = q.next;}//说明插在最后一个if(flag==0){q.next = temp;temp.next = null;}}return dummyNode.next;}
}

第二种，推荐面试直接写，堆排序，借助优先队列，能5分钟秒了

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode sortList(ListNode head) {//推荐使用优先队列PriorityQueue<ListNode> queue = new PriorityQueue<>(new Comparator<>(){public int compare(ListNode a ,ListNode b){return a.val-b.val;}});ListNode p = head;while(p!=null){queue.offer(p);p=p.next;}ListNode dummy = new ListNode(0);ListNode q = dummy;while(queue.isEmpty()==false){ListNode node = queue.poll();q.next = node;q = q.next;q.next = null;}return dummy.next;}
}

第三种，归并排序，主要就是分为归并和合并两部分。合并很简单。拆分时候，通过快慢指针拆出来两个链表头。

class Solution {public ListNode sortList(ListNode head) {if(head==null||head.next==null)return head;ListNode slow = head;//这里一定一定要slow和fast分开//不然 4 2 会出现，head一直是4，newHead一直是空的情况ListNode fast = head.next;while(fast!=null&&fast.next!=null){fast = fast.next.next;slow = slow.next;}ListNode newHead = slow.next;slow.next = null;return merge(sortList(head),sortList(newHead));}public ListNode merge(ListNode l1,ListNode l2){ListNode p = l1;ListNode q = l2;ListNode dummy = new ListNode(0);ListNode t = dummy;while(p!=null&&q!=null){if(p.val<q.val){t.next = p;p=p.next;}else{t.next = q;q=q.next;}t=t.next;}t.next = p==null?q:p;return dummy.next;}
}