費波那契數列由0和1開始,之後的數就由之前的兩數相加 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584,……….
用遞歸算法來求值,非常好理解.偽代碼:
f(n) = 0 (n=0)
f(n) = 1 (n=1)
f(n) = f(n-1) + f(n-2) (n>1)
實現:
def f(n):
if n==0:
return 0
elif n==1:
return 1
elif n>1:
return f(n-1) + f(n-2)
def f(n):
if n == 0:
return 0
if n == 1:
return 1
if n>1:
prev = 1 #第n-1項的值
p_prev = 0 #第n-2項的值
result = 1 #第n項的值
for i in range(1,n):
result = prev+p_prev
p_prev = prev
prev = result
return result
功能實現了,但是代碼比較冗長,函數是要對前兩項做特殊判斷.現在優化一下,如何才能更通用,即使是第0個和第1個也能運用到for循環呢?假設在 0, 1, 1, 2, 3, 5, 8, 13... 之前還有兩項, 是-1和1, 即: -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,這樣就通用了:
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
現在評估一下他們的性能: 寫一個性能裝飾器.
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
他不能用在遞歸方法中. 所以最終還是寫了這麼個方法:
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()
看出性能對比了吧:
54.2904469967
27.7642970085
所以用遞歸弊端還是不少