leetcode -- Remove Nth Node From End of List

Given a linked list, remove the n^th 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.


思路：两个指针，未能一次通过。。。。
当需删除的元素是第一个元素时，直接按24-27进行删除

/**
 * 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) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ListNode p = head;
        int totalNum = 0;
        while(p != null){
            totalNum ++;
            p = p.next;
        }
        
        
        if(totalNum == n){
25             head = head.next;
26             return head;
27         }
        
        
        ListNode p1 = head;
        for(int i = 0; i < (totalNum - n - 1); i ++){
            p1 = p1.next;
        }
        ListNode p2 = p1.next;
        if(p2 != null)
            p1.next = p2.next;
        else
            p1.next = null;
        
        return head;
        
    }
}