问题描述：

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.

思考：

要求给定的n总是有效的，且一趟遍历完成。两个指针，两个指针相隔n-1，也就是前面那个指针先跑n步，随后一起向后走，那么前面那个指针先跑到尾节点的时候，后面那个指针就指向被指定删除节点的前一个节点，把这个节点的next引用指向后面的后面的话就算完成了。

代码（java）：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        
        if(head == null){  
            return null;
        }
            
            
        ListNode front = head;
		ListNode back = head;		
		for(int i = 0 ; i < n ; i++){
			front = front.next;
		}
		
		
		if(front == null)  
        {  
            head = head.next;  
            front = null;  
            return head;  
        }
		
		while(front.next != null){
			front = front.next;
			back = back.next;
		}
		
		
		back.next = back.next.next;
		
		


		return head;
        
		
    }
}