题目描述：

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

解题思路一：递归的归并排序，两个关键点，找到中点mid和分割链表。前者通过快慢指针，后者通过指向None。即mid, slow.next = slow.next, None

然后对分割的链表再次进行递归分割，直到满足终止条件if not head or not head.next:
最后进行归并排序：空间复杂度：O(logn)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next: # termination.
            return head
        slow, fast = head, head.next
        while fast and fast.next:
            fast, slow = fast.next.next, slow.next
        mid, slow.next = slow.next, None # mid  奇数个节点找到中点右边，偶数个节点找到中心右边的节点。slow.next =  None 分割链表
        left, right = self.sortList(head), self.sortList(mid) # 递归分割
        h = res = ListNode(0)
        while left and right: # 归并排序
            if left.val < right.val:
                h.next, left = left, left.next
            else:
                h.next, right = right, right.next
            h = h.next
        h.next = left if left else right
        return res.next

时间复杂度：O(nlogn)
空间复杂度：O(logn)

解题思路二：归并排序（从底至顶直接合并）非递归

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        h, length, intv = head, 0, 1
        while h: h, length = h.next, length + 1
        res = ListNode(0)
        res.next = head
        # merge the list in different intv.
        while intv < length:
            pre, h = res, res.next
            while h:
                # get the two merge head `h1`, `h2`
                h1, i = h, intv
                while i and h: h, i = h.next, i - 1
                if i: break # no need to merge because the `h2` is None.
                h2, i = h, intv
                while i and h: h, i = h.next, i - 1
                c1, c2 = intv, intv - i # the `c2`: length of `h2` can be small than the `intv`.
                # merge the `h1` and `h2`.
                while c1 and c2:
                    if h1.val < h2.val: pre.next, h1, c1 = h1, h1.next, c1 - 1
                    else: pre.next, h2, c2 = h2, h2.next, c2 - 1
                    pre = pre.next
                pre.next = h1 if c1 else h2
                while c1 > 0 or c2 > 0: pre, c1, c2 = pre.next, c1 - 1, c2 - 1
                pre.next = h 
            intv *= 2
        return res.next

详见https://leetcode.cn/problems/sort-list/solutions/13728/sort-list-gui-bing-pai-xu-lian-biao-by-jyd
时间复杂度：O(nlogn)
空间复杂度：O(1)

解题思路三：0

时间复杂度：O(n)
空间复杂度：O(n)