【LeetCode】876. Middle of the Linked List 解题报告（Python）

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【LeetCode】876. Middle of the Linked List 解题报告（Python）

标签（空格分隔）： LeetCode

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.me/

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题目地址：https://leetcode.com/problems/middle-of-the-linked-list/description/

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.

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.

Note:

• The number of nodes in the given list will be between 1 and 100.

题目大意

给出链表中的中间节点。如果链表长度为偶数，返回的应该是中间的第二个节点。

解题方法

这个题很简单，就是快慢指针两个移动即可。

需要注意的是，如果链表长度为偶数，返回的应该是中间的第二个节点。这个做的方法，是判断fast指针是指向了最末尾的None还是链表结尾的node。如果是链表结尾说明是偶数个点，如果是none说明是奇数个点。

代码如下：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        dummy = ListNode(0)
        dummy.next = head
        slow, fast = dummy, dummy
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow.next if fast else slow

日期

2018 年 8 月 16 日 ———— 一个月不写题，竟然啥都不会了。。加油！