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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 };
LeetCode19 Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/wangxiaobao/p/5742904.html
