1ListNode* removeNthFromEnd(ListNode* head,intn) {2ListNode** t1 = &head, *t2 =head;3//t2向后移n个节点4while(n--) t2 = t2->next;5//使t2移到最后一个节点的next,即NULL6//t1指向那个指向待删除节点的指针,即指向待删除节点的上一个节点的next7while(t2 !=NULL) {8t2 = t2->next;9...
class Solution: # 定义一个方法,用于移除单向链表中的倒数第n个节点 def removeNthFromEnd(self, head, n): # 定义一个递归函数,用于实际执行移除操作 def remove(head): # 如果当前节点为空,则返回0和空节点。这表示已经到达链表末尾。 if not head: return 0, head # 递归调用remove函数处理当前节点的下...
Given a linked list, remove then-th node from the end of list and return its head. Example: Given linked list: 1->2->3->4->5, andn= 2. After removing the second node from the end, the linked list becomes 1->2->3->5. Note: Givennwill always be valid. Follow up: Could yo...
public ListNode removeNthFromEnd(ListNode head, int n) { // nth指向p前n个结点,pre为nth前一个结点,即pre/nth/p ListNode p = head, nth = head, pre = null; while (--n > 0) { p = p.next; } while (p.next != null) { p = p.next; pre = nth; nth = nth.next; } // 此...
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
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
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->...
简介: LeetCode 19. 删除链表的倒数第N个节点 Remove Nth Node From End of List LeetCode 19. 删除链表的倒数第N个节点 Remove Nth Node From End of ListTable of Contents一、中文版二、英文版三、My answer四、解题报告一、中文版给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。
var removeNthFromEnd = function(head, n) { let cur = head //迭代处理当前元素 let m = 0 //偏移量,用来指示要删除的元素 let del = head //要删除的元素 while (cur !== null) { if(m > n) { //达到并偏移指示窗口 del = del.next ...
Leetcode 19.Remove Nth Node From End of List 简介:删除单链表中的倒数第n个节点,链表中删除节点很简单,但这道题你得先知道要删除哪个节点。在我的解法中,我先采用计数的方式来确定删除第几个节点。另外我在头节点之前额外加了一个节点,这样是为了把删除头节点的特殊情况转换为一般情况,代码如下。