输入一个链表，输出该链表中倒数第k个结点。

'''输入一个链表，输出该链表中倒数第k个结点'''
#常规解法，考虑了代码的鲁棒性，考虑了空指针，K为0，K大于链表的长度
class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        if (head == None or k == 0):
            return None
        nodeFir = head
        nodeSco = head
        for i in range(1,k):
            if (nodeFir.next != None):
                nodeFir = nodeFir.next
            else:
                return None
        while(nodeFir.next != None):
            nodeFir = nodeFir.next
            nodeSco = nodeSco.next
        return nodeSco

#用栈的特性来解决这个问题，先压入栈，再取出K个值便是倒数第K个值
class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        if (head == None or k == 0):
            return None
        count = 0
        stack = []
        while (head != None):
            stack.append(head)
            head = head.next
            count += 1
            
        if (count < k):
            return None
        for i in range(0,k):
            kNode = stack.pop()
        return kNode