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...
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...
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...
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...
}// 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...
删除链表的倒数第N个节点 - 领扣 (LeetCode)leetcode-cn.com/problems/remove-nth-node-from-end-of-list/摘要: 本文适用于初学者。它介绍了以下内容: 链表的遍历和删除其末尾的第 n 个元素。方法一、两次遍历算法: 思路 我们注意到这个问题可以容易地简化成另一个问题:删除从列表开头数起的第 (L−n...
19. Remove Nth Node From End of List 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. ...
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
def removeNthFromEnd(self, head, n): """ :type head: ListNode :type n: int :rtype: ListNode """ my_head = my_trail = head for i in range(n): my_head = my_head.nextif not my_head: return head.nextwhile my_head.next: ...