中英题面
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。
Sort a linked list in O(n log n) time using constant space complexity.
示例 1:
输入: 4->2->1->3
输出: 1->2->3->4
Example 1:
Input: 4->2->1->3
Output: 1->2->3->4
示例 2:
输入: -1->5->3->4->0
输出: -1->0->3->4->5
Example 2:
Input: -1->5->3->4->0
Output: -1->0->3->4->5
算法
直接套用归并排序的思想，为了方便与节省空间，牺牲了一些常数时间。
时间复杂度：
O(NlogN)
空间复杂度：
O(1)
代码
1 # Definition for singly-linked list.
2 # class ListNode:
3 # def __init__(self, x):
4 # self.val = x
5 # self.next = None
6
7 class Solution:
8 def sortList(self, head):
9 """
10 :type head: ListNode
11 :rtype: ListNode
12 """
13 if (not head):
14 return None
15 if (not head.next):
16 return head
17 hen = tai = tail = head
18 while (tai.next):
19 tai = tai.next
20 if (tai.next):
21 tail = tail.next
22 else:
23 break
24 tai = tai.next
25 self.adjust(head, tai)
26 hen = tail.next
27 tail.next = None
28 self.adjust(head, tail)
29 self.adjust(hen, tai)
30 self.sortList(head)
31 self.sortList(hen)
32 i = head
33 while (hen):
34 if (i.next):
35 if (i.next.val > hen.val):
36 hen.next, i.next, hen = i.next, hen, hen.next
37 i = i.next
38 else:
39 i.next = hen
40 break
41 return head
42
43 def adjust(self, head, tail):
44 i = head
45 while (i.next):
46 if (head.val > i.next.val):
47 head.val, i.next.val = i.next.val, head.val
48 if (tail.val < i.next.val):
49 tail.val, i.next.val = i.next.val, tail.val
50 i = i.next