Remove Nth Node From End of List (Java)

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.

Note:
Given n will always be valid.
Try to do this in one pass.

用两个指针，一个指针先走n步，再与另一个指针一起走

Source

/**
 * 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) {
		if(head == null) return null;		//注意特殊情况 
		
		ListNode h1 = head;
		ListNode h2 = head;
		for(int i = 0; i < n; i++){
			 h2 = h2.next;
			
		}
		if(h2 == null){     //对应删除表头的情况
			head = head.next;
			return head;
		}
		while(h2.next != null){		//用h2的next来判断 这样方便删除
			h1 = h1.next;
			h2 = h2.next;
		}

		ListNode a;
		a = h1.next;
		h1.next = a.next;
		return head;		//返回表头即可
    }
}