148. Sort List [Medium] 归并排序

最新推荐文章于 2020-11-03 22:16:19 发布

木木歹朱

最新推荐文章于 2020-11-03 22:16:19 发布

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本文链接：https://blog.youkuaiyun.com/qq_35319215/article/details/96921196

博客围绕 148. Sort List 展开，要求时间复杂度为 O(nlogn)，采用归并排序。特别指出找 mid 时应让 fast = head.next，否则会出错。

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148. Sort List

要求时间复杂度O(nlogn)，归并排序，值得注意的是，找mid应该让fast = head.next，不然会出错。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head == None or head.next == None:
            return head
        mid = self.findMid(head)
        right = self.sortList(mid.next)
        mid.next = None
        left = self.sortList(head)
        return self.merge(left, right)
    def findMid(self, head):
        if head == None:
            return head
        fast, slow = head.next, head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        return slow
    def merge(self, l1, l2):
        if l1 == None:
            return l2
        if l2 == None:
            return l1
        dummy = p = ListNode(0)
        while l1 and l2:
            if l1.val < l2.val:
                p.next = l1
                l1 = l1.next
            else:
                p.next = l2
                l2 = l2.next
            p = p.next
        if l1:
            p.next = l1
        if l2:
            p.next = l2
        return dummy.next