LeetCode Remove Nth Node From End of List

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.

 

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode removeNthFromEnd(ListNode head, int n) {
14         ArrayList<ListNode> list=new ArrayList<>();
15         ListNode node=head;
16         int index=0;
17         while (node != null) {
18             list.add(node);
19             node=node.next;
20         }
21         index=list.size()-n;
22         if (index ==0) {
23             return head.next;
24         }
25         if (index==list.size()-1) {
26             list.get(index-1).next=null;
27             return head;
28         }
29         
30         list.get(index-1).next=list.get(index+1);
31         return head;
32         
33     }
34 }

 

posted @ 2014-10-24 11:46  birdhack  阅读(114)  评论(0)  编辑  收藏  举报