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...
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是为了让快慢指针之间保持n的距离# 同时让慢指针停在要删除节点的前一个节点for_inrange(n+1):fast...
}// 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...
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. 思路分析: 双指针,前后两个指针之间的距离为n-1,当后一个指针指向链表末尾结点的时候,前一个指针指向的结点就是要删除的结点。
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-node-from-end-of-list/solution/c-solution-by-shiminlei/算法 数据结构 leetcode 十月打卡挑战W3 ...
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
Try to do this in one pass. 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...