题目描述：Remove Nth Node From End of List
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 do this in one pass.
分析：
题目中说n是合法的，就不用对n进行检查了。用标尺的思想，两个指针相距为n-1，后一个到表尾，则前一个到n了。
① 指针p、q指向链表头部；
② 移动q，使p和q差n-1；
③ 同时移动p和q，使q到表尾；
④ 删除p。
(p为second，q为first)
代码如下：
/** * 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) { if(head == NULL || head->next == NULL) return NULL; ListNode * first = head; ListNode * second = head; for(int i = 0;i < n;i++) first = first->next; if(first == NULL) return head->next; while(first->next != NULL){ first = first->next; second = second->next; } second->next = second->next->next; return head; } };
LeetCode 019 Remove Nth Node From End of List
原文：http://www.cnblogs.com/510602159-Yano/p/4278912.html