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.

Solution:

 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         if (head==null) return head;
15 
16         ListNode first = head;
17         ListNode second = head;
18         for (int i=0;i<n;i++)
19             first = first.next;
20 
21         if (first==null) return second.next; 
22       
23         while (first.next!=null){
24             second = second.next;
25             first = first.next;
26         }
27 
28         ListNode temp = second.next.next;
29         second.next = temp;
30         return head;
31     }
32 }

 

posted @ 2014-11-29 10:26  LiBlog  阅读(110)  评论(0)  编辑  收藏  举报