LeetCode *876. Middle of the Linked List (Python Solution)

题目描述

Given a non-empty, singly linked list with head node head, return a middle node of linked list.

If there are two middle nodes, return the second middle node.

给定具有头节点头的非空的单链表，返回链表的中间节点。

如果有两个中间节点，则返回第二个中间节点。

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization:
[3,4,5]) The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans,
such that:
ans.val = 3, ans.next.val = 4,
ans.next.next.val = 5, and ans.next.next.next = NULL.

Example 2:

Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization:[4,5,6])
Since the list has two middle nodes with values 3 and 4, we
return the second one.

Python Solution

分析： 这道题在 LeetCode 中是 easy 的一道题。思维也很简单：1.先走一遍链表，计算长度。2.用math.ceil计算长度的一半取上值是多少。3.利用快慢指针进行计数找到要return的结点即可。

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
import math

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        length = 0
        count =head
        while count:
            length += 1
            count = count.next
        length = math.ceil(length/2)
        
        pre = post =head
        for _ in range(length):
            pre = pre.next
        while pre:
            pre = pre.next
            post = post.next
        return post

注意！

这里可能会有疑问：我们不是如果有两个中间结点的时候返回后面的结点么，那为什么用 math.ceil 向上取值，而不是用 math.floor 向下取值？
这里其实就是思路不清晰，如果有两个中间结点的时候肯定可以整除，而要用到 math 函数的时候其实是奇数个。

代码还可以简化成：

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        pre = post =head
        while pre and pre.next:
            pre = pre.next.next
            post = post.next
        return post

快指针走两步，慢指针走一步来达到走到中间结点的目的。要注意判断条件，必须确定快指针的下一个、下下个结点都不为None。