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.

链接：

4/11/2017

87%, 14ms

要了解每个变量的意义，想不出来就画图，不要猜初始值定在哪里

fast也可以设在dummy上，这样第8，12行就按照fast.next来判断，不过运行时间会加到18ms

1 public class Solution { 2     public ListNode removeNthFromEnd(ListNode head, int n) { 3         if (head == null || n <= 0) return head; 4         ListNode dummy = new ListNode(-1); 5         dummy.next = head; 6         ListNode fast = head; 7         ListNode slow = dummy; 8         while (n > 0 && fast != null) { 9             fast = fast.next;10             n--;11         }12         while (fast != null) {13             fast = fast.next;14             slow = slow.next;15         }16         slow.next = slow.next.next;17         return dummy.next;18     }19 }

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