[leetcode sort]148. Sort List

Sort a linked list in O(n log n) time using constant space complexity.

以时间复杂度O(n log n)排序一个链表。

归并排序，在链表中不需要多余的主存空间

tricks:用快慢指针找到中间位置，划分链表

 class Solution(object):
     def sortList(self, head):
         if not head or not head.next:
             return head
         pre, slow, fast = None, head, head
         while fast and fast.next:
             pre, slow, fast = slow, slow.next, fast.next.next
         pre.next = None
         return self.merge(*(map(self.sortList,(head,slow))))
     def merge(self,h1,h2):
         dummy = tail = ListNode(None)
         while h1 and h2:
             if h1.val < h2.val:
                 tail.next,h1, tail = h1,h1.next,h1
             else:
                 tail.next,h2,tail = h2,h2.next,h2
             tail.next = h1 or h2
         return dummy.next