# 排序链表

<p>给你链表的头结点 <code>head</code> ，请将其按 <strong>升序</strong> 排列并返回 <strong>排序后的链表</strong> 。</p>

<p><b>进阶：</b></p>

<ul>
	<li>你可以在 <code>O(n log n)</code> 时间复杂度和常数级空间复杂度下，对链表进行排序吗？</li>
</ul>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/f2b8957105894d4fe9ffe9462e65f8e8.jpeg" style="width: 302px; "/>
<pre>
<b>输入：</b>head = [4,2,1,3]
<b>输出：</b>[1,2,3,4]
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/f10bc169c916b5b0bc8651e5f7e99492.jpeg" style="width: 402px; " />
<pre>
<b>输入：</b>head = [-1,5,3,4,0]
<b>输出：</b>[-1,0,3,4,5]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<b>输入：</b>head = []
<b>输出：</b>[]
</pre>

<p> </p>

<p><b>提示：</b></p>

<ul>
	<li>链表中节点的数目在范围 <code>[0, 5 * 10<sup>4</sup>]</code> 内</li>
	<li><code>-10<sup>5</sup> <= Node.val <= 10<sup>5</sup></code></li>
</ul>


## template

```python
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        if head == None:
            return None
        else:
            return self.mergeSort(head)

    def mergeSort(self, head):
        if head.next == None:
            return head
        fast = head
        slow = head
        pre = None

        while fast != None and fast.next != None:
            pre = slow
            slow = slow.next
            fast = fast.next.next
        pre.next = None

        left = self.mergeSort(head)
        right = self.mergeSort(slow)

        return self.merge(left, right)

    def merge(self, left, right):
        tempHead = ListNode(0)
        cur = tempHead
        while left != None and right != None:
            if left.val <= right.val:
                cur.next = left
                cur = cur.next
                left = left.next
            else:
                cur.next = right
                cur = cur.next
                right = right.next
        if left != None:
            cur.next = left
        if right != None:
            cur.next = right
        return tempHead.next

```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```