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.

 

 这道题要注意的Corner Case是：如果n比这个LinkedList的size大，那么就需要直接返回Head，如果相等，就把head删掉，返回head.next；基本做的方法呢，还是Runner Technique. 由于有可能head会被删掉，所以最好还是使用一个dummy node，dummy.next = head。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode removeNthFromEnd(ListNode head, int n) {
11         ListNode dummy = new ListNode(-1);
12         dummy.next = head;
13         ListNode p1 = dummy, p2 = dummy;
14         while (p2.next != null) {
15             if (n > 0) {
16                 n --;
17             }
18             else p1 = p1.next;
19             p2 = p2.next;
20         }
21         if (n == 0) p1.next = p1.next.next;
22         return dummy.next;
23     }
24 }