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
https://www.nowcoder.com/practice/54404a78aec1435a81150f15f899417d
import sys
def do(l_num, k):
return l_num[-k]
if __name__ == '__main__':
l_lines = sys.stdin.readlines()
for index in range(len(l_lines)):
if index % 3 == 1:
l_num = [int(i) for i in l_lines[index].strip().split(' ')]
k = int(l_lines[index+1].strip())
r = do(l_num, k)
print(r)