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;
	}
}