利用Python实现斐波那契数列的5种方法全解析

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

def fib_recursive(n):
    if n <= 1:
        return n
    return fib_recursive(n - 1) + fib_recursive(n - 2)

from functools import lru_cache
 
@lru_cache(maxsize=None)
def fib_memo(n):
    if n <= 1:
        return n
    return fib_memo(n - 1) + fib_memo(n - 2)

def fib_generator(n):
    a, b = 0, 1
    count = 0
    while count < n:
        yield a
        a, b = b, a + b
        count += 1
 
# 使用方式
for num in fib_generator(10):
    print(num, end=' ')
# 输出：0 1 1 2 3 5 8 13 21 34

def matrix_multiply(A, B):
    return [
        [A[0][0]*B[0][0] + A[0][1]*B[1][0], A[0][0]*B[0][1] + A[0][1]*B[1][1]],
        [A[1][0]*B[0][0] + A[1][1]*B[1][0], A[1][0]*B[0][1] + A[1][1]*B[1][1]]
    ]
 
def matrix_power(mat, n):
    if n == 1:
        return mat
    if n % 2 == 0:
        half = matrix_power(mat, n // 2)
        return matrix_multiply(half, half)
    else:
        return matrix_multiply(mat, matrix_power(mat, n - 1))
 
def fib_fast(n):
    if n <= 1:
        return n
    base_matrix = [[1, 1], [1, 0]]
    result_matrix = matrix_power(base_matrix, n)
    return result_matrix[0][1]

"""
【推荐】斐波那契函数合集（可直接复制使用）
"""
from functools import lru_cache
 
# 1. 循环迭代（最优）
def fib_iter(n):
    a, b = 0, 1
    for _ in range(n):
        a, b = b, a + b
    return a
 
# 2. 记忆化递归（推荐学习）
@lru_cache(maxsize=None)
def fib_memo(n):
    if n <= 1:
        return n
    return fib_memo(n - 1) + fib_memo(n - 2)
 
# 3. 生成器（内存友好）
def fib_gen(n):
    a, b = 0, 1
    for _ in range(n):
        yield a
        a, b = b, a + b
 
# 测试
if __name__ == "__main__":
    print("前10个斐波那契数：")
    print(list(fib_gen(10)))