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. 示例: 给定一个链表: 1->2->3->4->5, 和 n = 2. 当删除了倒数第二个节点后,链表变为 1->2->3->5. 说明: 给定的 n 保证是有效的. 进阶: 你能尝试使用一趟扫描实现吗? 方法一: 两次遍历,第一次求长度. 方法二: 一次遍历, first指针和second指针中间隔了n个节点,second ->next就是要删除的节点.当要删除的倒数长度和链表的长度相同时…

题目: 删除链表中倒数第n个节点 给定一个链表,删除链表中倒数第n个节点,返回链表的头节点.  样例 给出链表1->2->3->4->5->null和 n = 2. 删除倒数第二个节点之后,这个链表将变成1->2->3->5->null. 注意 链表中的节点个数大于等于n 解题: 要删除倒数第n个节点,我们要找到其前面一个节点,也就是倒数第n+1的节点,找到这个节点就可以进行删除.和上题的思想很类似, 定义两个指针,p和cur,cur指针向前走,走了n…

Given a linked list, remove the n-th node from the end of list and return its head. 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 w…

给定一个链表,删除链表的倒数第 n 个节点并返回头结点.例如,给定一个链表: 1->2->3->4->5, 并且 n = 2.当删除了倒数第二个节点后链表变成了 1->2->3->5.详见:https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/ 思路:设置两个指针first跟second.first指针先移动n步,若此时first指针为空,则表示要删除的是头节点,此时直…

这道题是LeetCode里的第19道题. 题目要求: 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. 示例: 给定一个链表: 1->2->3->4->5, 和 n = 2. 当删除了倒数第二个节点后,链表变为 1->2->3->5. 说明: 给定的 n 保证是有效的. 进阶: 你能尝试使用一趟扫描实现吗? 先求链表长度的暴力法我就不讲了. 关于一趟实现的办法就是先让 back 先走 n 次,此时 front 和 back 产生了长度为 n 的间…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:链表, 删除节点,双指针,题解,leetcode, 力扣,Python, C++, Java 目录 题目描述 题目大意 解题方法 双指针 日期 题目地址:https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/ 题目描述 Given a linked list,…

题目意思:去掉链表中倒数第n个节点 思路:1.两次遍历,没什么技术含量,第一次遍历计算长度,第二次遍历找到倒数第k个,代码不写了   2.一次遍历,两个指针,用指针间的距离去计算. ps:特别注意删掉第一个节点的情况,在前面加个头结点,则变为了删除第二个节点的情况 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x),…

/** * 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, int n) { ,*w=,*e=; ; ) return head; //无需修改…

Given a linked list, remove the n-th node from the end of list and return its head. 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 w…

Given a linked list, remove the n-th node from the end of list and return its head. 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 w…

题目:删除链表的倒数第N个节点 难度:Medium 题目内容: Given a linked list, remove the n-th node from the end of list and return its head. 翻译:给定一个链表,删除倒数第n个节点并返回其头部. Example: Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, t…

Level:   Medium 题目描述: Given a linked list, remove the n-th node from the end of list and return its head. 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-…

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. Given a linked list, remove the n-th node from the end of list and return its head. 示例: 给定一个链表: 1->2->3->4->5, 和 n = 2. 当删除了倒数第二个节点后,链表变为 1->2->3->5. 说明: 给定的 n 保证是有效的. Note: Given n will always be val…

题目描述: 中文: 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. 示例: 给定一个链表: 1->2->3->4->5, 和 n = 2. 当删除了倒数第二个节点后,链表变为 1->2->3->5. 说明: 给定的 n 保证是有效的. 进阶: 你能尝试使用一趟扫描实现吗? 英文: Given a linked list, remove the n-th node from the end of list and return its head.…

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…

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…

题目 Given a linked list, remove the n-th node from the end of list and return its head. *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: Give…

视频讲解  http://v.youku.com/v_show/id_XMTY1MTMzNjAyNA==.html (1)定义两个指针 ListNode fast = head; ListNode slow = head; (2)将快指针向前移动N步 (3.1)判断此时快指针是否已经到达尽头,如果是,头节点就是要删除的节点,返回head.next. (3.2)将快慢两个指针同时以相同的速度往前移动,当快指针走到尽头的时候,慢指针的下一个位置就是倒数第N个节点,将慢指针next指向next.nex…

https://leetcode.com/problems/remove-nth-node-from-end-of-list/ 使用双指针法,可以仅遍历一次完成节点的定位 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public:…

题目描述 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. 示例: 给定一个链表: 1->2->3->4->5, 和 n = 2. 当删除了倒数第二个节点后,链表变为 1->2->3->5. 说明: 给定的 n 保证是有效的. 进阶: 你能尝试使用一趟扫描实现吗? 解题思路 典型的利用双指针法解题.首先让指针first指向头节点,然后让其向后移动n步,接着让指针sec指向头结点,并和first一起向后移动.当first的next指针为NULL时,…

题意:删除链表中倒数第N个节点. 法一:递归.每次统计当前链表长度,如果等于N,则return head -> next,即删除倒数第N个节点:否则的话,问题转化为子问题“对head->next这个链表删除倒数第N个节点”,将head的next指针指向该子问题的结果,返回head即可.这个方法时间复杂度高,不推荐. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next;…

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ 方法: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 + 1)(L−n+1) 个结点,其中 LL 是列表的长度.只要我们找到列表的长度 LL,这个问题就很容易解决. 算法 首先我们…

原题地址:http://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/ 题意: 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 n…

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…

Swap Nodes in Pairs Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the va…

Remove Nth Node From End of List Total Accepted: 46720 Total Submissions: 168596My Submissions Question Solution  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,…

1.合并两个排好序的list Merge Two Sorted Lists Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 2.删除list倒数第n个元素 Remove Nth Node From End of List Given a linked list,…

leetcode-algorithms-19 Remove Nth Node From End of List Given a linked list, remove the n-th node from the end of list and return its head. Example: Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the…

61. Rotate List(M) Given a list, rotate the list to the right by k places, where k is non-negative. For example: Given ->->->->->NULL and k = , ->->->->->NULL. Total Accepted: 102574 Total Submissions: 423333 Difficulty: Medi…

我现在在做一个叫<leetbook>的开源书项目,把解题思路都同步更新到github上了,需要的同学可以去看看 这个是书的地址: https://hk029.gitbooks.io/leetbook/ 19. Remove Nth Node From End of List 问题 Given a linked list, remove the nth node from the end of list and return its head. For example, Given linked…