Remove Nth Node From End of List

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.

删除倒数第N个节点。设两个分开n距离的指针 同时前进  当后一个指针指向null时 前一个指针就到要删除的节点啦  然后删除这个节点。（考虑特殊情况 删除头尾节点）

/** * 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||head.next==null)       return null;       ListNode l1=head;       ListNode l2=head;       ListNode l3=l1;       while(n>0){           l2=l2.next;           n--;       }       while(l2!=null){           l3=l1;//注意 这里用l3保留l1前一个节点  方便最后节点的删除。也可以用编程之美里面删除任意节点的方法           l1=l1.next;           l2=l2.next;       }       if(l1==head){           head=head.next;       }       if(l1.next==null){           l3.next=null;       }       else{       l3.next=l1.next;       }       return head;    }}