费波那契数列由0和1开始，之后的数就由之前的两数相加 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584,……….
递归算法
用递归算法来求值,非常好理解.伪代码:
f(n) = 0 (n=0)
f(n) = 1 (n=1)
f(n) = f(n-1) + f(n-2) (n>1)
实现:
def f(n):
if n==0:
return 0
elif n==1:
return 1
elif n>1:
return f(n-1) + f(n-2)
非递归算法
def f(n):
if n == 0:
return 0
if n == 1:
return 1
if n>1:
prev = 1 #第n-1项的值
p_prev = 0 #第n-2项的值
result = 1 #第n项的值
for i in range(1,n):
result = prev+p_prev
p_prev = prev
prev = result
return result
功能实现了,但是代码比较冗长,函数是要对前两项做特殊判断.现在优化一下,如何才能更通用,即使是第0个和第1个也能运用到for循环呢?假设在 0, 1, 1, 2, 3, 5, 8, 13... 之前还有两项, 是-1和1, 即: -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,这样就通用了:
def f(n):
prev = 1
p_prev = -1
result = 0
for i in range(n+1):
result = prev+p_prev
p_prev = prev
prev = result
return result
现在评估一下他们的性能: 写一个性能装饰器.
def perfromce_profile(func):
def wrapper(*args, **kwargs):
start = time.time()
rtn = func(*args, **kwargs)
end = time.time()
print end-start
return rtn
return wrapper
他不能用在递归方法中. 所以最终还是写了这么个方法:
import time
def f0(n):
if n==0:
return 0
elif n==1:
return 1
elif n>1:
return f(n-1) + f(n-2)
def f(n):
prev = 1
p_prev = -1
result = 0
for i in range(n+1):
result = prev+p_prev
p_prev = prev
prev = result
return result
def perfromce_profile():
start = time.time()
f0(1000000)
end = time.time()
print end-start
start = time.time()
f(1000000)
print time.time()-start
if __name__ == '__main__':
perfromce_profile()
看出性能对比了吧:
54.2904469967
27.7642970085
所以用递归弊端还是不少
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