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.

解法：三個指標，p1指向尾節點，p2指向倒數第K個節點，pre指向要刪除節點的前一個節點，注意特殊情況，剛好要刪除的節點是頭結點的情況

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode p1 = head;
        ListNode p2 = null, pre=null;
        int k = 0;
        
        while(p1 != null){
            k++;
            p1 = p1.next;
            
            if(k == n){
                p2 = head;
            } 
            else if(k > n){
                pre = p2;
                p2 = p2.next;
            }
        }
        
        if(p2 != null){
            if(pre == null) head = p2.next;
            else pre.next = p2.next;
        }
        
        return head;
    }
}