Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

1. 需要一个dummy node，dummy位置是0，prev在dummy
2. 先让head走 从1 开始i <= n，跳出循环head在n＋1上，head和prev相隔n个
3. 同时走，head到null的时候，prev正好在要删除的node的前一个

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;

        for(int i = 1; i <= n; i++ ){
            head = head.next;
        }
        while (head != null){
            head = head.next;
            pre = pre.next;
        }
        pre.next = pre.next.next;
        return dummy.next;

    }
}