#19 Remove Nth Node From End of List

题目：

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.

题解：

还是Easy的难度。要求one pass，two pointers可以很好解决：left和right相差n，当right到结尾的时候，left就是要去掉的链表节点。另外需要记录要去掉的节点的前一节点。

特殊情况有一种：linked list 只有一个节点，且n=1.

我没注意的情形：去掉的是头结点的情况！！！

/** * 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.next==null)return null;        ListNode left=head;        ListNode right=head;        ListNode pre=null;        int diff=0;        while(right!=null){            if(diff<n){                right=right.next;                diff++;            }            else{                right=right.next;                pre=left;                left=left.next;            }        }        if(left==head)head=head.next;//这句开始没想到，Runtime error        else pre.next=left.next;        return head;    }}

PS：我加上那句中文注释再提交，运行时间从280ms直接到460ms，orz。