# [排序链表](https://leetcode-cn.com/explore/interview/card/bytedance/244/linked-list-and-tree/1040/)

**头条重点**

## 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
```

## 解题思路

  1. 通过快慢指针将链表拆分
  2. 递归进行拆分，再通过合并两个排序链表的方式进行合并
  3. 类似于归并排序

```
public ListNode sortList(ListNode head) {
    if (head == null || head.next == null) {
        return head;
    }

    ListNode slow = head, fast = head;
    while (fast.next != null && fast.next.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }

    ListNode mid = slow.next;
    slow.next = null;

    ListNode l1 = sortList(head);
    ListNode l2 = sortList(mid);

    return merge(l1, l2);
}

private ListNode merge(ListNode l1, ListNode l2) {
    if (l1 == null) {
        return l2;
    }

    if (l2 == null) {
        return l1;
    }

    ListNode head,res;
    if (l1.val > l2.val) {
        head = l2;
        l2 = l2.next;
    } else {
        head = l1;
        l1 = l1.next;
    }
    res = head;
//        head.next = null;

    while (l1 != null || l2 != null) {
        if (l1 == null) {
            head.next = l2;
            l2 = l2.next;
        } else if (l2 == null) {
            head.next = l1;
            l1 = l1.next;
        } else {
            if (l1.val > l2.val) {
                head.next = l2;
                l2 = l2.next;
            } else {
                head.next = l1;
                l1 = l1.next;
            }
        }
        head = head.next;
    }

    return res;
}
```
