Remove Nth Node From End of List

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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.

按照双指针的思想去考虑这个问题。比如给出了一个链表 1 2 3 4 5 , 如果想要去除导数第二个元素，让两个指针都指向头，先让一个指针走两步到达3，然后两个指针一起向右移动，当右指针的next为空值时，删除左指针的next 即可。上面的例子中，左指针指向3，右指针指向5。此时，先暂时保留左指针的next.next的位置，然后赋值给左指针.next ，将元素4删去。提交的时候会出现一个元素的情况，还有空元素的情况，都需要考虑到。

代码：

package leetcode;public class RemoveN { public static void main(String[] args) {   RemoveN r = new RemoveN();        // TODO Auto-generated method stub          ListNode head = new ListNode(1);          ListNode cur = head;          for(int i = 1; i < 5; i++)          {                         ListNode tmp = new ListNode(i+1);              cur.next = tmp;              cur = tmp;          }                ListNode n1 = new ListNode(1);        ListNode n2 = new ListNode(2);        n1.next = n2;              cur = r.removeNthFromEnd(n1,2);                    for(;cur != null;)          {              System.out.println(cur.val);              cur = cur.next;          }                      }   public ListNode removeNthFromEnd(ListNode head, int n) { if(head == null )               return head;  if(head.next ==null){ return null; }        //双指针思想 ListNode p1 = head; ListNode p2 = head; for(int i=0;i<n;i++){if(p2==null){//说明给的n大了，return  null;}p2 = p2.next;} if (p2 == null){ //删除头结点 head = head.next; return head;  }else{  while(p2.next != null){ p1= p1.next; p2= p2.next; } //p2到达后，删除对应的p1结点 ListNode tmp = p1.next.next; p1.next=tmp; return head; }}}      class ListNode {      int val;      ListNode next;      ListNode(int x) { val = x; }  }

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