leetcode 148 Sort List（python

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描述

Given the head of a linked list, return the list after sorting it in ascending order.Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?

Example 1:

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ini复制代码Input: head = [4,2,1,3]
Output: [1,2,3,4]

Example 2:

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ini复制代码Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]

Example 3:

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ini复制代码Input: head = []
Output: []

Note:

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kotlin复制代码The number of nodes in the list is in the range [0, 5 * 10^4].
-10^5 <= Node.val <= 10^5

解析

根据题意，就是给出了一个链表的头节点 head ，要求我们对其进行升序排序。题目还要求我们使用 O(n logn) 的时间复杂度和 O(1) 的内存。我尝试用了最简单的暴力插入，空间复杂度还是 O(1)，但是 O(n^2) 时间复杂度，还是超时了。

解答

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python复制代码class ListNode(object):
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:return head
        L = ListNode(val=-1000000)
        L.next  = cur = head
        while cur.next:
            pre = L
            if cur.next.val >= cur.val:
                cur = cur.next
                continue
            while pre.next.val < cur.next.val:
                pre = pre.next
            tmp  = cur.next
            cur.next = tmp.next
            tmp.next = pre.next
            pre.next = tmp
        return L.next

运行结果

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css复制代码Time Limit Exceeded

解析

因为这里用到了内置函数 sorted 对值进行排序，又重新建立新的链表，所以时间上复杂度上是 O(n logn) 但是竟然通过了汗颜。但是空间复杂度是 O(n)。

解答

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python复制代码class ListNode(object):
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        node = ListNode(val = 0)
        result = node
        nlist = []
        while head != None:
            nlist.append(head.val)
            head = head.next
        nlist = sorted(nlist)
        for n in nlist:
            node.next = ListNode(val = n)
            node = node.next
        return result.next

运行结果

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erlang复制代码Runtime: 284 ms, faster than 84.91% of Python online submissions for Sort List.
Memory Usage: 63.3 MB, less than 12.52% of Python online submissions for Sort List.

解析

还可以用归并排序，只要是时间复杂度为 O(n logn) 的其他排序算法都可以。

解答

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python复制代码class ListNode(object):
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:return head
        mid = self.getMid(head)
        another = mid.next
        mid.next = None
        return self.merge(self.sortList(head), self.sortList(another))
    
    def getMid(self, head):
        fast = slow = head
        while fast.next and fast.next.next:
            fast = fast.next.next
            slow = slow.next
        return slow
    
    def merge(self, A, B):
        dummy = cur = ListNode(0)
        while A and B:
            if A.val > B.val:
                cur.next = B
                B = B.next
            else:
                cur.next = A
                A = A.next
            cur = cur.next
        if A: cur.next = A
        if B: cur.next = B
        return dummy.next

运行结果

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erlang复制代码Runtime: 544 ms, faster than 44.25% of Python online submissions for Sort List.
Memory Usage: 45.6 MB, less than 86.45% of Python online submissions for Sort List.

原题链接：leetcode.com/problems/so…

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