单向链表倒数第 k 个结点(Python)
输出单向链表中倒数第k个结点
http://www.nowcoder.com/questionTerminal/54404a78aec1435a81150f15f899417d
解法一:
装作没看到,当数组解 :)
while True:
try:
l, s, k = int(input()), input().split(), int(input())
print(s[l - k] if k else 0)
except:
break 解法二:
自己定义节点再连成链表咯 :)
class Node(object):
def __init__(self, val=0):
self.val = val
self.next = None
while True:
try:
l, s, k, head = int(input()), list(map(int, input().split())), int(input()), Node()
while k:
head.next = Node(s.pop())
head = head.next
k -= 1
print(head.val)
except:
break 提交结果
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