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.

解题思路:1.利用双指针，先让快指针跳N个结点，然后两个指针一起往后直到快指针为NULL，此时慢的指针指向结点即为所求结点

               2.利用哈希的原理，将结点按顺序存在vector，利用vector索引找到所求结点.

#include<iostream>
#include<vector>
using namespace std;

//Definition for singly - linked list.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};
//解法一:双指针
ListNode *removeNthFromEnd(ListNode *head, int n) {
	ListNode*FastN_node = head;
	for (int i = 0; i != n; ++i)
		FastN_node = FastN_node->next;
	ListNode*SlowN_node = head;
	ListNode*PreSlowN_node = SlowN_node;
	while (FastN_node!=NULL)
	{
		PreSlowN_node = SlowN_node;
		FastN_node    = FastN_node->next;
		SlowN_node    = SlowN_node->next;
	}
	if (SlowN_node == head)
		head = head->next;
	else
		PreSlowN_node->next = SlowN_node->next;
	return head;
}
//解法二:哈希
ListNode *removeNthFromEnd(ListNode *head, int n)
{
	vector<ListNode*>NodeListVector;
	ListNode* TmpHead = head;
	int TotalNum = 0;
	for (; TmpHead != NULL;TmpHead=TmpHead->next,++TotalNum)
		NodeListVector.push_back(TmpHead);
	if (TotalNum - n - 1 < 0)
		head = head->next;
	else
		NodeListVector[TotalNum - n - 1]->next = NodeListVector[TotalNum - n]->next;
	return head;
}

Remove Nth Node From End of List