题目

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 。
示例 1：
输入：head = [4,2,1,3]
输出：[1,2,3,4]

分析

这道题目的思路也比较明确，我们可以用递归排序的思路，先不断拆分链表直到只剩下一个结点，然后再合并俩个有序列表。拆分链表可以用快慢指针的方法

public class LinkNode {int val;LinkNode next;public LinkNode(int data) {this.val = data;this.next = null;}
}
public class LinkList {LinkNode head;public LinkList() {this.head = null;}public LinkNode getHead() {return this.head;}//添加元素public void addNode(int data) {LinkNode node = new LinkNode(data);if (this.head == null) {this.head = node;} else {LinkNode cur = this.head;while(cur.next != null) {cur = cur.next;}cur.next = node;}}//正序打印public void print(LinkNode node) {while(node != null) {System.out.print(node.val);System.out.print(" ");node = node.next;}System.out.println();}public LinkNode sort(LinkNode head) {if(head == null || head.next == null) {return head;}LinkNode fast = head;LinkNode slow = head;LinkNode pre = null;while(fast != null && fast.next != null) {pre = slow;slow = slow.next;fast = fast.next.next;}pre.next = null;return merge(sort(slow),sort(head));}public LinkNode merge(LinkNode slow,LinkNode fast) {LinkNode dummy = new LinkNode(0);LinkNode p = dummy;while(slow != null && fast != null) {if(slow.val < fast.val) {p.next = slow;p = p.next;slow = slow.next;} else {p.next = fast;p = p.next;fast = fast.next;}}while(slow != null) {p.next = slow;p = p.next;slow = slow.next;}while(fast != null) {p.next = fast;p = p.next;fast = fast.next;}return dummy.next;}
}
public class sortList {public static void main(String[] args) {LinkList list = new LinkList();list.addNode(-1);list.addNode(5);list.addNode(3);list.addNode(4);list.addNode(0);LinkNode head = list.sort(list.getHead());list.print(head);}
}