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T9-斐波那契用迭代

跳台阶

动规 py2

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloor(self, number):
        # write code here
        if number<0:
            return -1
        if number <=2:
            return number
        a,b=1,2
        res = 0
        for i in range(3,number+1):
            res = a+b
            a,b = b,res
        return res

变态跳台阶

wo的初始迭代方法

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloorII(self, number):
        # write code here
        if number<0:
            return -1
        res = [0,1]
        for i in range(2,number+1):
            res.append(sum(res)+1)
        return res[number]

数学--移位 py2

调到第n级台阶，前面的(n-1)级有跳与不跳两种选择吗，故答案为2^(n-1),并且用移位运算代替乘法运算来优化。

# -*- coding:utf-8 -*-
class Solution:
    def jumpFloorII(self, number):
        # write code here
        if number<=0:
            return -1
        res = 1
        return res<<(number-1)

矩形覆盖

动规 Py2

# -*- coding:utf-8 -*-
class Solution:
    def rectCover(self, number):
        # write code here
        if number<0:
            return -1
        if number<=2:
            return number
        a,b=1,2
        res=0
        for i in range(3,number+1):
            res = a+b
            a ,b = b,res
        return res

剑指offer 斐波那契系列

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原文地址：https://www.cnblogs.com/yxl-2018/p/12395396.html