还是老老实实直接写代码。 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...
"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...
Given a linked list, remove the nth node from the end of list and return its head. For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5. Note: Given n will always be valid. Try to d...
Leetcode: Remove Nth Node From End of List 题目: Given a linked list, remove the nth node from the end of list and return its head. For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->...
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After removing the second node from the end, the linked list becomes 1->2->3->5. Note: Given n will always be valid. Try to do this in one pass. 移除链表中倒数第n个节点; 写一个函数来确定要删除的是第几个节点,方法是获取链表总长度len,那么我们要删除的就是第 nth = len - n 个节点...
19. Remove Nth Node From End of List 题意 给一个链表和一个数n, 删掉倒数第n个数. 比如给出1->2->3->4->5和2 返回1->2->3->5 尽量保证只遍历一次数组(one-pass) 解法: 常规思路是先完整遍历一遍数组,得到数组的长度L. 然后再从头遍历一次, 删除第L-n个数. ...
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
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