148. 排序链表

1. 题目链接

2. 解题思路：在链表上使用排序算法。注意，不能使用快排，因为快排的最差时间复杂度是O(n^2)，数组形式的快排，以随机数划分能够得到O(n*logn)，但是链表的形式，不太好以随机数的方式划分。所以最好的排序方法是使用归并排序。

   • 先用快慢指针，将链表分成两部分，然后两部分分别归并排序，得到两个链表的头和尾。然后再merge即可

3. 代码

   class Solution:
    
    
       def process(self, node: Optional[ListNode]) -> (Optional[ListNode], Optional[ListNode]):
           slow = node
           if slow.next == None:
               return slow, slow
           fast = slow.next
           while fast and fast.next:
               slow = slow.next
               fast = fast.next
               if fast == None:
                   break
               fast = fast.next
           # 分别有序
           tmp = slow.next
           slow.next = None
           head1, tail1 = self.process(node)
           head2, tail2 = self.process(tmp)
           # 合并
           ans_head  = None
           ans_tail = tail1 if tail1.val >= tail2.val else tail2
           if head1.val <= head2.val:
               ans_head = head1
               head1 = head1.next
           else:
               ans_head = head2
               head2 = head2.next
           cur = ans_head
           while head1 and head2:
               if head1.val <= head2.val:
                   cur.next = head1
                   head1 = head1.next
               else:
                   cur.next = head2
                   head2 = head2.next
               cur = cur.next
           if head1:
               cur.next = head1
           if head2:
               cur.next = head2
    
           return ans_head, ans_tail
    
    
    
       def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
           if head == None:
               return None
           head1, _ = self.process(head)
           return head1