地址：https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

Given a linked list, remove the n^th 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.

其实题目要求遍历链表只能是一次，然后不需要考虑n 值的合法性。最简单办法就是先遍历求得链表节点数目，然后删除指定结点，但是这样需要两次遍历链表。

public class Solution {       public ListNode removeNthFromEnd(ListNode head, int n) {    	if(head == null){    		return null;    	}    	ListNode countList = head;    	int ListCount = 0;    	while(countList!=null){    		ListCount++;    		countList = countList.next;    	}    	if(ListCount-n==0){    		return head.next;    	}    	ListNode ans = head;    	int count = 0;    	ListNode pre = null;    	ListNode cur = null;        while(head.next!=null){        	count++;        	pre = head;        	cur = head.next;        	if(count == ListCount-n){        		pre.next = cur.next;        		break;        	}        	head = head.next;        }        return ans;    }}

有个方法可以只需要进行一次链表遍历，参见：http://blog.csdn.net/huruzun/article/details/22047381

public class Solution {        public ListNode removeNthFromEnd(ListNode head, int n){    	ListNode pre = null;    	ListNode ahead = head;    	ListNode behind = head;    	// ahead 节点先走n-1 步，behind 和ahead 一起走，当ahead到达最后一个节点，behind为需要删除元素    	for(int i=0;i<n-1;i++){    		if(ahead.next!=null){    			ahead = ahead.next;    		}else {				return null;			}    	}    	while(ahead.next!=null){    		pre = behind;    		behind = behind.next;    		ahead = ahead.next;    	}    	// 如果删除的是头结点    	if(pre == null){    		return behind.next;    	}else {			pre.next = behind.next;		}    	return head;    }}

个人觉得第二种方法也没有什么太多优化，只是为了满足题目要求。但是这个方法联想到可以解决很多链表相关问题，比如求两个链表相交结点也可以通过这种类似这种路程关系解决。

Remove Nth Node From End of List