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 node from the end, the linked list becomes 1->2->3->5. Try to do this in one pass. (Easy)

分析：题目虽然是easy，但是用one pass做起来还是包含几个链表常用的手法的，有参考价值；

链表算法上没有什么难的，就是代码上仔细，再熟悉几种处理方法即可。

1.Two pointers。 利用快慢指针，快指针先走n步，然后一起走，快的下一步到终点了，慢的到达要删除的前一位；

2. dummy node。 凡是head位置可能被删掉，返回出现问题，用dummy node处理返回问题。

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 dummy(0);13         dummy.next = head;14         head = &dummy;15         ListNode* chaser = head;16         ListNode* runner = head;17         for (int i = 0; i < n; ++i) {18             runner = runner -> next;19         }20         while (runner -> next != nullptr) {21             runner = runner -> next;22             chaser = chaser -> next;23         }24         ListNode* temp = chaser -> next;25         chaser -> next = chaser -> next -> next;26         delete temp;27         return dummy.next;28     }29 };