【leetcode】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.

解析：删除一串链表中倒数第n个节点，不用考虑n非法的情况（比如n大于链表的长度），题目要求以一次遍历解决。可以用“快指针和慢指针”解决问题，将快指针先走n个节点，然后两个指针再已同样的步调同时向末尾移动，当快指针为null时，慢指针为待删除节点。

需要注意的是，题目要求最后返回链表的头结点，因为头结点也可以能是待删除节点，所以可以用一个“哑节点”dummy做标记，dummy.next即表示当前的头结点。

Java AC代码：

public class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode dummy = new ListNode(-1);dummy.next = head;ListNode fast = dummy;ListNode slow = dummy;while (n-- > 0) {fast = fast.next;}while(true){if(fast.next == null){slow.next = slow.next.next;break;}fast = fast.next;slow = slow.next;}return dummy.next;}}