输入一个链表，输出该链表中倒数第k个结点。

* solution1: Two pointers
```python3
# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        if not head or k <=0:
            return None
        p2 = head
        p1 = head
        while k > 1:
            if p2.next:
                p2 = p2.next
                k -= 1
            else:
                return None
        while p2.next:
            p1 = p1.next
            p2 = p2.next
        return p1

```

* solution2: Brute Force
```python3
# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        res = []
        while head:
            res.append(head)
            head = head.next
        if len(res) < k or k <1:
            return
        return res[-k]
```