链表--Leetcode 876 Middle of the Linked List

链表--Leetcode 876 Middle of the Linked List"/>

链表--Leetcode 876 Middle of the Linked List

链表–Leetcode 876 Middle of the Linked List

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.

来源：力扣（LeetCode）
链接：
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思路解析

• 解法一：根据链表的长度，来获得中间结点(粗糙)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def middleNode(self, head: ListNode) -> ListNode:if not head or not head.next:return headlength = self.get_length(head)for i in range(0, length // 2):head = head.nextreturn headdef get_length(self, head):length = 0temp = headwhile temp:length += 1temp = temp.nextreturn length

时间复杂度 O ( N ) O(N) O(N)，空间复杂度 O ( 1 ) O(1) O(1)

• 解法二：快慢指针

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def middleNode(self, head: ListNode) -> ListNode:if not head or not head.next:return headslow = headfast = headwhile fast and fast.next:slow = slow.nextfast = fast.next.nextreturn slow 

时间复杂度 O ( N ) O(N) O(N)，空间复杂度 O ( 1 ) O(1) O(1)

注意偶数长度和奇数长度的区别

• 解法三，将链表存放到列表中，利用列表的索引获取中间元素

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def middleNode(self, head: ListNode) -> ListNode:if not head or not head.next:return headres = [head]while res[-1].next:res.append(res[-1].next)return res[len(res) // 2]

时间复杂度 O ( N ) O(N) O(N)，空间复杂度为 O ( N ) O(N) O(N)