【Leetcode Algorithm】Remove Nth Node From End of List

题目：

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.（意思是O(n)复杂度）

刚开始只想到用两个指针，中间隔n-1步，当走在前面的指针到达尾部时，那么后一个节点一定指向倒数第n个节点。这样还得记录倒数第n个节点的前驱节点，从而才能删除倒数第n个节点。但是，这样遇到的问题是，如果指向的该倒数第n个节点是head节点（没有前驱节点了），那么就没法删除了。改进的方法就是，从一开始就新建一个节点behindhead，使其指向头结点。最后返回的时候返回behindhead.next就好了

代码：

/**
 * 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) {
        //新建一个节点指向head
        ListNode behindhead = new ListNode(0);
        behindhead.next = head;
        //建两个指针，指针ahead一个指向head节点
        ListNode ahead = head;
        //指针behind指向behindhead节点
        ListNode behind = behindhead;
        //指针ahead需要向前移动n-1步
        for(int i=0; i<n-1; i++){
            if(ahead.next!=null){
                ahead = ahead.next;
            }
        }
        //当指针ahead没有到达尾部时，指针ahead和behind都往前移动
        while(ahead.next!=null){
            ahead = ahead.next;
            behind = behind.next;
        }
        //删除behind后面的节点
        behind.next = behind.next.next;
        return behindhead.next;
    }
}