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...
* };*/classSolution {public: ListNode* removeNthFromEnd(ListNode* head,intn) {if(!head)returnNULL; stack<ListNode*>s; ListNode* node =head;while(node){ s.push(node); node= node->next; }for(inti =0; i < n; ++i) { s.pop(); }if(s.empty()) {returnhead->next; }else{ s....
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...
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 * */ for(int ...
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->...
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
训练营的课程视频是免费的都在b站,但是资源群(指导刷题顺序,分享刷题经验,美国面试经验)是收费的。收费目的是刷掉打广告的,浑水摸鱼的等不是真心想来刷题的人。QQ群号:623125309 。如果无法加入参见我的置顶动态。题解:https://leetcode-cn.com/problems/remove-nth-
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. ...
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