【LeetCode】Remove Nth Node From End of List

题目描述：

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.

第一种方法可以先遍历一遍统计下总长度，再算下前序长度，不过这样显然没太有意思。

第二种是递归，从后往前计算长度，也差不多吧……需要注意下移出第一节点时的处理。

因为if后面的==写成了=，没有one pass，真是想哭……

class Solution {
public:
	ListNode *next;
	ListNode *removeNthFromEnd(ListNode *head, int n) {
		next = NULL;
		if (recusive(head, n) == n)
			return head->next;
		return head;
	}
	int recusive(ListNode *node, int n){
		int num = 1;
		if (node->next)
			num += recusive(node->next, n);
		if (num == n - 1)
			next = node;
		if (num == n + 1)
			node->next = next;
		return num;
	}
};