LeetCode: Remove Nth Node From End of List 解题报告

Remove Nth Node From End of List

 Total Accepted: 46720 Total Submissions: 168596My Submissions

 

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.

 

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SOLUTION 1:

1.使用快慢指针,快指针先行移动N步。用慢指针指向要移除的Node的前一个Node.

2. 使用dummy node作为head的前缀节点,这样就算是删除head也能轻松handle啦!

主页君是不是很聪明呀? :)

 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         // 0017
15         ListNode dummy = new ListNode(0);
16         dummy.next = head;
17         
18         ListNode slow = dummy;
19         ListNode fast = dummy;
20         
21         // move fast N more than slow.
22         while (n > 0) {
23             fast = fast.next;
24             // Bug 1: FORGET THE N--;
25             n--;
26         }
27         
28         while (fast.next != null) {
29             fast = fast.next;
30             slow = slow.next;
31         }
32         
33         // Slow is the pre node of the node which we want to delete.
34         slow.next = slow.next.next;
35         
36         return dummy.next;
37     }
38 }
View Code

 

GITHUB (国内用户可能无法连接):

https://github.com/yuzhangcmu/LeetCode/blob/251766ffb832f2278f43a05e194ca76584bf14ea/list/RemoveNthFromEnd.java

posted on 2015-04-04 15:25  Yu's Garden  阅读(778)  评论(0编辑  收藏

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