关于LeetCode中Remove Nth Node From End of List一题的理解

题目如下：

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.

    题目要求删除倒数第n个链表元素，通过设置一个静态变量，然后使用递归我们很轻易地就能解决。唯一可能出现的问题在于LeetCode的judger可能对含有静态变量的函数出现误判的情况，具体情况可以看这里：https://leetcode.com/faq/#different-output。解决方案就是加一个默认构造器即可，关于java的默认构造器可以看看这篇文章：http://www.2cto.com/kf/201311/255432.html。好了，直接上已Accepted的代码：

public class Solution {
static int count;
public Solution(){
count=1;
}
public ListNode removeNthFromEnd(ListNode head, int n) {
if(head==null){
return null;
}
head.next = removeNthFromEnd(head.next,n);
if(count==n){
head = head.next;
}
count++;
return head;
}
}    这道题是有Editorial Solution的，Editorial Solution的牛逼之处在于它没有使用静态变量，地址在这里，里面附有详细的解释：https://leetcode.com/articles/remove-nth-node-end-list/。

    我这里先把它的两种方法代码放上去，具体解释可以看上面链接，第一种方法代码如下：

public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
int length = 0;
ListNode first = head;
while (first != null) {
length++;
first = first.next;
}
length -= n;
first = dummy;
while (length > 0) {
length--;
first = first.next;
}
first.next = first.next.next;
return dummy.next;
}
    第二种方法代码如下：

public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode first = dummy;
ListNode second = dummy;
// Advances first pointer so that the gap between first and second is n nodes apart
for (int i = 1; i <= n + 1; i++) {
first = first.next;
}
// Move first to the end, maintaining the gap
while (first != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}