【leetcode】sort list（python）

链表的归并排序

超时的代码

class Solution:
	def merge(self, head1, head2):
		if head1 == None:
			return head2
		if head2 == None:
			return head1
		# head1 and head2 point to the same link list
		if head1 == head2:
			return head1

		head = None
		tail = None
		# the small pointer point to smaller of two.
		while head1 and head2:
			if head1.val <= head2.val:
				small = head1
				head1 = head1.next
			else:
				small = head2
				head2 = head2.next
			# the first node
			if tail == None:
				tail = small
				head = small
			else:
				tail.next = small
				tail = small
		# link the remaind nodes
		if head1 == None:
			head1 = head2

		tail.next = head1
		return head

	def sortList(self, head):
		if head == None or head.next == None:
			return head
		# we use a fast pointer which go two steps each time and
		# a slow pointer which go one step each time to get the
		# middle of the link list
		slow = head
		fast = head
		while fast.next and fast.next.next:
			fast = fast.next.next
			slow = slow.next
		# slow point to the middle now
		head2 = slow.next
		# we cut of the linked list at middle
		slow.next = None

		left = self.sortList(head)
		right = self.sortList(head2)
		return self.merge(left, right)

主要是在merge的时候。要推断第一个结点

AC代码

class Solution:
	def merge(self, head1, head2):
		if head1 == None:
			return head2
		if head2 == None:
			return head1
		# head1 and head2 point to the same link list
		if head1 == head2:
			return head1

		head = ListNode(-1)
		tail = head
		# the small pointer point to smaller of two.
		while head1 and head2:
			if head1.val <= head2.val:
				small = head1
				head1 = head1.next
			else:
				small = head2
				head2 = head2.next

			tail.next = small
			tail = small
		# link the remaind nodes
		if head1 == None:
			head1 = head2

		tail.next = head1
		return head.next

	def sortList(self, head):
		if head == None or head.next == None:
			return head
		# we use a fast pointer which go two steps each time and
		# a slow pointer which go one step each time to get the
		# middle of the link list
		slow = head
		fast = head
		while fast.next and fast.next.next:
			fast = fast.next.next
			slow = slow.next
		# slow point to the middle now
		head2 = slow.next
		# we cut of the linked list at middle
		slow.next = None

		left = self.sortList(head)
		right = self.sortList(head2)
		return self.merge(left, right)

这里merge的时候建立了一个伪头结点，处理的时候就不用推断是否为第一个结点，尽管AC，但时间5500ms以上。而c++代码仅仅须要200多ms，差距还是比較大的