删除单链表的倒数第n个元素

题目原型：

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.

这个题目比较简单，就不分析了，直接贴代码。

public ListNode removeNthFromEnd(ListNode head, int n) {//求出链表长度int len = 0;//链表长度ListNode p,q;p = head;while(p!=null){len++;p = p.next;}if(n>len)return null;p = head;q = head;for(int i = 0;i<len-n;i++){p = q;q = q.next;}//当删除的是第一个节点的时候if(p==q)head = p.next;if(q!=null){p.next = q.next;q.next = null;}return head;    }