1.用循环打印出斐波那契数列。
2.用递归打印出斐波那契对应的数字。
用循环制作
def rabbit2(n):
i = 2
x1 = 0
x2 = 1
x3 = 0
a = [1]
if n >= 2:
while i < n + 1:
x3 = x1 + x2
x1 = x2
x2 = x3
i += 1
a.append(x3)
else:
return a
else:
return a
n = 8
d = rabbit2(n)
print(d)
n = 1
d = rabbit2(n)
print(d)
用递归制作
第一版：重复计算版
def rabbit1(n):
if n <= 1:
return n
else:
return rabbit1(n - 1) + rabbit1(n - 2)
print('Pls print a number')
print（rabbit1（4））
print（rabbit1（12）
第二版
a = 0
rab = [0]*100
def rabbit2(n):
global a
a += 1
if n <= 0:
return 0
elif n == 1:
return 1
else:
b = rabbit2(n - 1)
rab[n - 1] = b
c = rab[n - 2]
return b + c
print(rabbit2(100), '{} : a'.format(a))
print(rabbit2(50),'{} : a'.format(a))
用递归+数列制作，更方便：
用递归+列表制作
def rabbit1(n) :
rab = []
if n <= 1:
return 1
else:
rab.append(1)
rab.append(1)
for i in range(2,n):
rab.append(rab[i - 1] + rab[i - 2])
return rab
print('Pls print a number')
print(rabbit3(1000))
用class解决该问题
class Rabbit:
def __init__(self, cap):
self.rab = [0] * cap
self.c = 0
def count(self, n):
self.c += 1
if n <= 0:
return 0
elif n == 1:
return 1
else:
if self.rab[n - 1] > 0:
a = self.rab[n - 1]
else:
a = self.count(n - 1)
self.rab[n - 1] = a
if self.rab[n - 2] > 0:
b = self.rab[n - 2]
else:
b = self.count(n - 2)
self.rab[n - 2] = b
return a + b
修改后的class
class Rabbit:
def __init__(self, cap):
self.rab = [0] * cap
self.c = 0
def count(self, n):
self.c += 1
if n <= 0:
return 0
elif n == 1:
return 1
else:
a = self.count(n - 1)
self.rab[n - 1] = a
b = self.rab[n - 2]
return a + b
