算法描述：
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->5.
Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?
解题思路：
这个题考查数据结构。设置两个指针，快指针先走n步，然后快指针和慢指针一起走，快指针停的时候，慢指针的下一个元素为待删除元素。
ListNode* removeNthFromEnd(ListNode* head, int n) { ListNode* dup = new ListNode(-1); dup->next = head; ListNode* fast = dup; while(n>0){ fast=fast->next; n--; } ListNode* slow = dup; while(fast!=nullptr && fast->next!=nullptr){ fast=fast->next; slow=slow->next; } slow->next = slow->next->next; return dup->next; }
LeetCode-19-Remove Nth Node From End of List
原文：https://www.cnblogs.com/nobodywang/p/10315115.html