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题目描述：
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;
}
}
Java [leetcode 19]Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/zihaowang/p/4492590.html
