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
# “机智”解法
# 一
import sys
lines = sys.stdin.readlines()
lines = sys.stdin.readlines()
for i in range(len(lines)//3):
n,values,k = int(lines[3*i]),lines[3*i+1].split(),int(lines[3*i+2])
n,values,k = int(lines[3*i]),lines[3*i+1].split(),int(lines[3*i+2])
print(values[-k])
# 二
class Node:
def __init__(self,value=0):
self.value=value
self.next=None
import sys
lines = sys.stdin.readlines()
for i in range(len(lines)//3):
n,vs,k = int(lines[3*i]),lines[3*i+1].split(),int(lines[3*i+2])
node=None
while vs and k>0:
node=Node(vs.pop())
k-=1
print(node.value)
class Node:
def __init__(self,value=0):
self.value=value
self.next=None
import sys
lines = sys.stdin.readlines()
for i in range(len(lines)//3):
n,vs,k = int(lines[3*i]),lines[3*i+1].split(),int(lines[3*i+2])
node=None
while vs and k>0:
node=Node(vs.pop())
k-=1
print(node.value)
