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...
【LeetCode】26.Linked List —Remove Nth Node From End of List 从列表末尾删除第n个节点 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...
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...
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...
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...
Remove Nth Node From End of List 2. Solution Version 1 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution{public:ListNode*removeNthFromEnd(ListNode*head,intn){if(!head||...
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n的元素,可以在第二个元素后面跟一...
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