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Leetcode-Remove Nth Node From End of List

时间:2014-11-29 11:44:24      阅读:127      评论:0      收藏:0      [点我收藏+]

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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 }

 

Leetcode-Remove Nth Node From End of List

标签:style   blog   io   color   sp   for   strong   on   div   

原文地址:http://www.cnblogs.com/lishiblog/p/4130201.html

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