* Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */classSolution{publicListNodemiddleNode(ListNode head){ListNodefast=head, slow = head;while(fast !=null&& fast.next !=null) { slow = slow.next; fast ...
Given the head of a singly linked list, return the middle node of the linked list.If there are two middle nodes, return the second middle node. 英文版地址 leetcode.com/problems/m 中文版描述 给定一个头结点为 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。
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:Node3fromthis list (Serialization: [3,4,5]) The returned node has value3. (The j...
/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */ func middleNode(head *ListNode) *ListNode { // 初始化快慢指针均为头结点 fast := head slow := head // 如果快指针还能走两步,则继续循环 for fast != nil && fast.Next != nil ...
876. Middle of the Linked List 题目分析 返回一个链表中最中间的元素。 思路 先全部塞入数组,再根据长度/2得到中间元素的下标,再返回。 最终代码 <?php /** * Definition for a singly-linked list. * class ListNode { * public $val = 0; ...
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....
给你单链表的头结点head,请你找出并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:head = [1,2,3,4,5]输出:[3,4,5]解释:链表只有一个中间结点,值为 3 。 示例2: 输入:head = [1,2,3,4,5,6]输出:[4,5,6]解释:该链表有两个中间结点,值分别为 3 和 4 ...
// 链表的中间结点(LeetCode 876):https://leetcode.cn/problems/middle-of-the-linked-list/ classSolution{ publicListNodemiddleNode(ListNode head){ // 设置两个指针,一开始都指向链表的头节点 ListNode slow = head; ListNode fast = head; // 接下来,让这两个指针向前移动 ...
struct ListNode*middleNode(struct ListNode*head){struct ListNode*slower=head;struct ListNode*faster=head;//注意边界条件while((faster!=NULL)&&(faster->next!=NULL)){faster=faster->next->next;slower=slower->next;}returnslower;} 328. 奇偶链表 ...
* ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */ class Solution { public: ListNode* middleNode(ListNode* head) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2 head = [1,2,3,4,5] 1 2 [1,2,3,4,5] [1,2,3,4,5,6] Source ...