题解 | #求二叉树的层序遍历#
跳台阶
http://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4
# -*- coding:utf-8 -*-
class Solution:
def jumpFloor(self, number):
n2 = n
return n
class Solution:
def jumpFloor(self, number):
# write code here
# 初始化n1,n2为1层和2层时的方法数
n1 = 1
n2 = 2
# 当层数为0时,返回0
if number == 0:
return 0
# 当层数为1时,有1种方法
if number == 1:
return n1
当层数为2时,有2种方法
if number == 2:
return n2
# 当为3层以上时
for i in range(3, number+1):
# 到达本层的方法数为:上一层和上两层的方法数之和
n = n1 + n2
# 传递更新n1和n2
n1 = n2n2 = n
return n