[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