题解 | #输出单向链表中倒数第k个结点#
输出单向链表中倒数第k个结点
http://www.nowcoder.com/practice/54404a78aec1435a81150f15f899417d
因为华为认为倒数0节点上是没有值的，所以注意处理k是0的情况
无脑list啊，纯考试够用
while True:
try:
count, num_list, k = int(input()), [int(x) for x in input().split()], int(input())
print(num_list[-k] if k else 0)
except EOFError:
break
链表做法，直接逆向遍历
class Node(object):
def __init__(self, val=0):
self.val = val
self.next = None
while True:
try:
head = Node()
count, num_list, k = int(input()), list(map(int, input().split())), int(input())
while k:
head.next = Node(num_list.pop())
head = head.next
k -= 1
print(head.val)
except EOFError:
break
