Java [leetcode 19]Remove Nth Node From End of List

题目描述：

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.

解题思路：

设置两个指针，两个指针相隔n-1，然后两个指针同时向后移动，当后一个指针没有后继节点了，那么前一个指针指向的节点就是需要删除的节点。

代码如下：

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) { val = x; }

 * }

 */

public class Solution {

    public ListNode removeNthFromEnd(ListNode head, int n) {

		if(head == null)

			return null;

		ListNode pPre = null;

		ListNode p = head;

		ListNode q = head;

		for(int i = 0; i < n - 1; i++)

			q = q.next;

		while(q.next != null){

			pPre = p;

			p = p.next;

			q = q.next;

		}

		if(pPre == null)

			return head.next;

		pPre.next = p.next;

		return head;

	}

}