Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

题目链接：https://leetcode.com/problems/remove-nth-node-from-end-of-list/

题目分析：fir先前进n，然后和sec一起前进，注意判断删除的是表头的情况

5ms，时间击败100%

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode fir = head;
        ListNode sec = head;
        while (fir.next != null) {
            if (n > 0) {
                n--;
            } else {
                sec = sec.next;   
            }
            fir = fir.next;
        }
        if (n > 0) {
            return head.next;
        }
        if (sec.next != null) {
            sec.next = sec.next.next;
        }
        return head;
    }
}