[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:

快慢指针。快的比慢的快n个位置，这样快的指针指到末尾的时候，慢的指针刚好距离list末尾n个位置。

 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         ListNode dummy=new ListNode(0);
15         dummy.next=head;
16         
17         ListNode fast=dummy, slow=dummy;
18         while(fast!=null&&n>0){
19             fast=fast.next;
20             n--;
21         }
22         while(fast.next!=null){
23             fast=fast.next;
24             slow=slow.next;
25         }
26         slow.next=slow.next.next;
27         return dummy.next;
28     }
29 }

第一次提交的时候报错，问题在：

slow.next=slow.next.next;

我写成了

slow.next=fast;

当时由于自己只在草稿纸上画了下n=2的例子，而忘记了n的含义：nth node from the end of list.