LeetCode | 0876. Middle of the Linked List链表的中间结点【Python】

LeetCode 0876. Middle of the Linked List链表的中间结点【Easy】【Python】【双指针】

Problem

LeetCode

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.

问题

力扣

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。

思路

快慢指针

快指针每次移动两个位置,慢指针每次移动一个位置。
当快指针到链表尾的时候,慢指针正好到中间位置。

时间复杂度: O(n),n 为链表长度。
空间复杂度: O(1)

Python3代码
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        if not head:
            return None
        
        # 快慢指针
        slow, fast = head, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow

GitHub链接

Python

posted @ 2020-03-23 22:39  Wonz  阅读(147)  评论(0编辑  收藏  举报