Fibonacci

斐波那契数列(Fibonacci)递归与非递归的性能对比

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 timedef 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()-startif __name__ == '__main__':    perfromce_profile() 

看出性能对比了吧:

54.290446996727.7642970085

所以用递归弊端还是不少