题目：
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.
给定一个单链表，移除倒数第n个Node，返回链表的head
思路：维护两个指针slow，fast，fast先移动n步，然后slow、fast同时移动，直到fast为空为止，删除slow后面的结点即可
代码：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* removeNthFromEnd(ListNode* head, int n) { 12 ListNode *preHead = new ListNode(0); 13 ListNode *fast = preHead, *slow = preHead; 14 slow->next = head; 15 for(int i = 0; i <= n; ++i) 16 fast = fast->next; 17 while(fast) 18 { 19 fast = fast->next; 20 slow = slow->next; 21 } 22 slow->next = slow->next->next; 23 return preHead->next; 24 } 25 };
[LeetCode19]Remove Nth Node From End of List
原文：http://www.cnblogs.com/zhangbaochong/p/5186060.html