题目:移除linked-list从尾到头的第N个元素 自我思路:因为题目给出的N是从链表尾开始计算的,单链表不存在从子指向父亲的反向指针,因此我先计算链表的整个长度len,然后用len - N来表示正向被删除元素所在的位置. 代码: public ListNode removeNthFromEnd(ListNode head, int n) { if(head == null) return null; ListNode help = head; int length = 1; while(he…

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…

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,…

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 …

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…

https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/ remove倒数第n个节点 一般list remove node的题目,都要先设置一个 dummy 节点, dummy->next = head,最后返回 dummy->next. 因为有可能要删除的就是head节点,这样不用再额外判断了. /** * Definition for singly-linked list. * struct ListNode {…

Given a linked list, remove the n th 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:  Give…

# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def removeNthFromEnd(self, head, n): """ :type head: ListNode :type n: int :rtype: ListNode &q…