Fibonacci数Python的四种解法

# 计算Fibonacci数：
# Naive版本，时间效率O(1.618^n)
# 记忆化版本(增加line8、10、13)，时间效率O(n)
# 注意：当n超过1000，可能超过系统允许的最大递归深度

from time import process_time

# memo = {}
def fib(n): 
    # if n in memo: return memo[n]
    if n < 2: f = n
    else: f = fib(n-2) + fib(n-1)
    # memo[n] = f
    return f

n = int(input(‘n = ‘))
start = process_time()
print(fib(n))
elapsed = (process_time() - start)
print("用时（秒）:",elapsed)

# 计算Fibonacci数：
# 迭代版本，时间效率O(n)，空间效率O(n)
# 注意：当n超过1000000，可能出现Memory Error

from time import process_time

memo = {}
def fib(n):
    for i in range(n+1):
        if i < 2: f = i
        else: f = memo[i-1]+memo[i-2]
        memo[i] = f
    return memo[n]

n = int(input(‘n = ‘)) 
start = process_time()
print(fib(n))
elapsed = (process_time() - start)
print("用时（秒）:",elapsed)

# 计算Fibonacci数：
# 空间效率优化为O(2)，时间效率O(n)

from time import process_time

def fib(n):
    a, b = 0, 1
    for i in range(n):
        a, b = b, a + b
    return a

n = int(input(‘n = ‘)) 
start = process_time()
print(fib(n))
elapsed = (process_time() - start)
print("用时（秒）:",elapsed)

# 计算Fibonacci数：
# 利用矩阵幂次求解，由于使用分治方法，时间效率O(log(n))

from time import process_time

def fib(n): 
      
    F = [[1, 1], 
         [1, 0]] 
    if (n == 0): 
        return 0
    power(F, n - 1) 
          
    return F[0][0] 
      
def multiply(F, M): 
      
    x = (F[0][0] * M[0][0] + 
         F[0][1] * M[1][0]) 
    y = (F[0][0] * M[0][1] + 
         F[0][1] * M[1][1]) 
    z = (F[1][0] * M[0][0] + 
         F[1][1] * M[1][0]) 
    w = (F[1][0] * M[0][1] + 
         F[1][1] * M[1][1]) 
      
    F[0][0] = x 
    F[0][1] = y 
    F[1][0] = z 
    F[1][1] = w 
          
def power(F, n): 
  
    if( n == 0 or n == 1): 
        return; 
    M = [[1, 1], 
         [1, 0]]; 
          
    power(F, n // 2) 
    multiply(F, F) 
          
    if (n % 2 != 0): 
        multiply(F, M) 

n = int(input(‘n = ‘)) 
start = process_time()
print(fib(n))
elapsed = (process_time() - start)
print("用时（秒）:",elapsed)