LeetCode:Remove Nth Node From End of List

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.

一开始的想法是遍历一遍得到LinkedList的长度，在得到要删除的位置，在删除。

后来看到Note中提到最好只遍历一次。后来的想法是，在一次遍历过程中，在i位置上查看其n个位置后的Node是不是终止结点。如果是则i位置为要删除结点删除就好。

public class Solution {

     public ListNode removeNthFromEnd(ListNode head, int n)

    {

        ListNode start = head;

        ListNode former = head;

        while(!isNextNthNodeEndOfList(start,n))

        {

            former = start;

            start = start.next;

        }

        if(start == former)return start.next;

        former.next = start.next;

        return head;

    }

   
    public boolean isNextNthNodeEndOfList(ListNode node,int n)

    {

        int i = 1;

        ListNode p = node;

        while(i<n)

        {

            p = p.next;

            i++;

        }

        if(p.next == null) return true;

        return false;

       
    }

}