"null" : next.value) + '}'; } } public static void main(String[] args) { RemoveNthNodeFromEndOfList removeNthNodeFromEndOfList = new RemoveNthNodeFromEndOfList(); Node head = new Node(); Node last = head; last.value = 1; for (int i = 2; i <= 5; i++) { Node node ...
}// 删除第n个节点// ListNode nthNode = p1.next;p1.next = p1.next.next;// nthNode.next = null;returndummy.next; } }// Runtime: 6 ms// Your runtime beats 100.00 % of java submissions. Python 实现 # Definition for singly-linked list.# class ListNode:# def __init__(self, x...
classSolution:defremoveNthFromEnd(self,head,n):# 初始化一个哑节点,它的下一个节点指向链表头节点# 这样做是为了方便处理边界情况,比如删除的是头节点dummy=ListNode(0)dummy.next=head# 初始化快慢指针,初始时都指向哑节点slow=dummyfast=dummy# 快指针先前进n+1步,走到第n+1个节点# 这里加1是为了让快慢...
Can you solve this real interview question? Remove Nth Node From End of List - Given the head of a linked list, remove the nth node from the end of the list and return its head. Example 1: [https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.j
public ListNode removeNthFromEnd11(ListNode head, int n) { if(head==null || n<=0) return head; ListNode start=head,end=head; /* * 因为n是有效的,所以这个循环不会报错,最多end为null走到了最后 * 这里是循环次数为n的for循环,表示的子链表的间隔是n,也就是说元素是n+1个 ...
Given the head of a linked list, remove the nth node from the end of the list and return its head.impl Solution { pub fn remove_nth_from_end(head: Option<Box<ListNode>>, n: i32) -> Option<Box<ListNode>> { let mut dummy = Some(B
【Leetcode】Remove Nth Node From End of List https://leetcode.com/problems/remove-nth-node-from-end-of-list/ 题目: nth For example, AI检测代码解析 Given linkedlist:1->2->3->4->5,andn=2.After removing the second node from the end,the linkedlistbecomes1->2->3->5....
public ListNode removeNthFromEnd3(ListNode head, int n) { ListNode temp = new ListNode(0, head); int length = 0; while (head != null) { length++; head = head.next; } ListNode cur = temp; for (int i = 1; i < length - n + 1; ++i) { cur = cur.next; } cur.next = ...
After removing the second node from the end, the linked list becomes 1->2->3->5. Note: Given n will always be valid. Follow up: Could you do this in one pass? 链表操作的题,比较简单,主要主要Bug Free public ListNode removeNthFromEnd(ListNode head, int n) { ...
原问题链接:https://leetcode.com/problems/remove-nth-node-from-end-of-list/ 问题分析 这个问题相对来说比较简单。因为要删除链表中从后往前的第n个元素,所以需要用一个元素first的引用先往前n步,然后再用一个引用second跟着这个元素一步步到链表的最后。既然要删除这个倒数第n的元素,可以在第二个元素后面跟一...