python递归与非递归实现斐波那契数列
1.题目描述
大家都知道斐波那契数列，现在要求输入一个整数n，请你输出斐波那契数列的第n项（从0开始，第0项为0）。
递归实现：
class Solution():
def Fibnacci(self,n):
if n <= 0:
return 0
if n == 1:
return 1
return self.Fibnacci(n-1) + self.Fibnacci(n-2)
非递归实现：
def Fibnacci(n):
result = [0,1]
if n <= 1:
return result[n]
for i in range(2,n+1):
result.append(result[i-1]+result[i-2])
return result[n]
2.题目描述
一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法（先后次序不同算不同的结果）。
一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法（先后次序不同算不同的结果）。
[2. Tímù miáoshù
yī zhǐ qīngwā yīcì kěyǐ tiào shàng 1 jí táijiē, yě kěyǐ tiào shàng 2 jí. Qiú gāi qīngwā tiào shàng yīgè n jí de táijiē zǒnggòng yǒu duōshǎo zhǒng tiào fǎ (xiānhòu cìxù bùtóng suàn bùtóng de jiéguǒ).]
yī zhǐ qīngwā yīcì kěyǐ tiào shàng 1 jí táijiē, yě kěyǐ tiào shàng 2 jí. Qiú gāi qīngwā tiào shàng yīgè n jí de táijiē zǒnggòng yǒu duōshǎo zhǒng tiào fǎ (xiānhòu cìxù bùtóng suàn bùtóng de jiéguǒ).]
2. Description of the topic
A frog can jump to the first step or jump to the second level. Ask the frog to jump on an n-level step in total how many kinds of jumps (the order is different).
A frog can jump to the first step or jump to the second level. Ask the frog to jump on an n-level step in total how many kinds of jumps (the order is different).
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