# 排序链表

## 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]
示例 2：


输入：head = [-1,5,3,4,0]
输出：[-1,0,3,4,5]
示例 3：

输入：head = []
输出：[]
 

提示：

链表中节点的数目在范围 [0, 5 * 104] 内
-105 <= Node.val <= 105
 

进阶：你可以在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？

## 思路

- 归并排序
  - 拆分
    - 快慢指针
  - 合并


## 代码

```java
/**
 * 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) {
        // -- Divide -> Sort -> Merge --
        // Special
        if (head == null || head.next == null) {
            return head;
        }

        // Init: slow, fast, dummyHead
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode slow = dummyHead;
        ListNode fast = dummyHead;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        // Divide: [headA .... tailA(slow)] [headB(slow.next) ... tailB]
        ListNode headA = head;
        ListNode headB = slow.next;
        slow.next = null;
        // Sort
        headA = sortList(headA);
        headB = sortList(headB);
        // Merge
        return mergeList(headA, headB);
    }

    public ListNode mergeList(ListNode headA, ListNode headB) {
        // Init: dummyHead, tail
        ListNode dummyHead = new ListNode(-1);
        ListNode tail = dummyHead;
        // Traverse: Merge
        while (headA != null && headB != null) {
            int valA = headA.val;
            int valB = headB.val;
            if (valA < valB) {
                tail.next = headA;
                headA = headA.next;
            } else {
                tail.next = headB;
                headB = headB.next;
            }
            tail = tail.next;
        }
        // Operation: After
        tail.next = headA == null ? headB : headA;
        return dummyHead.next;
    }
}
```