LeetCode 876. 链表的中间结点
Table of Contents
一、中文版
给定一个带有头结点 head 的非空单链表,返回链表的中间结点。
如果有两个中间结点,则返回第二个中间结点。
示例 1:
输入:[1,2,3,4,5]
输出:此列表中的结点 3 (序列化形式:[3,4,5])
返回的结点值为 3 。 (测评系统对该结点序列化表述是 [3,4,5])。
注意,我们返回了一个 ListNode 类型的对象 ans,这样:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, 以及 ans.next.next.next = NULL.
示例 2:
输入:[1,2,3,4,5,6]
输出:此列表中的结点 4 (序列化形式:[4,5,6])
由于该列表有两个中间结点,值分别为 3 和 4,我们返回第二个结点。
提示:
给定链表的结点数介于 1 和 100 之间。
二、英文版
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. 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/middle-of-the-linked-list 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
三、My answer
class Solution: def middleNode(self, head: ListNode) -> ListNode: fast = head slow = head while fast is not None and fast.next is not None: # while fast and fast.next: fast = fast.next.next slow = slow.next return slow
四、解题报告
快慢指针,都从头节点开始移动。快指针每次移动两步,慢指针每次移动一步,当 fast 是空或者 fast.next 是空时,此时慢指针所指位置就是链表的中间节点。