[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。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy=new ListNode(0);
        dummy.next=head;
        
        ListNode fast=dummy, slow=dummy;
        while(fast!=null&&n>0){
            fast=fast.next;
            n--;
        }
        while(fast.next!=null){
            fast=fast.next;
            slow=slow.next;
        }
        slow.next=slow.next.next;
        return dummy.next;
    }
}

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.