python斐波那契数列的计算方法

#!usr/bin/python2.7
# -*- coding=utf8 -*-
# @Time  : 18-1-3 下午2:53
# @Author : Cecil Charlie

import sys
import copy
sys.setrecursionlimit(1000) # 用来调整解释器默认最大递归深度


class Fibonacci(object):
  def __init__(self):
    pass

  def fibonacci1(self, n):
    '''
      原始的方法，时间复杂度为 o(2**n)，因此代价较大
    :param n: 数列的第n个索引
    :return: 索引n对应的值
    '''
    if n < 1:
      return 0
    if n == 1 or n == 2:
      return 1
    return self.fibonacci1(n-1) + self.fibonacci1(n-2)

  @staticmethod
  def fibonacci2(n):
    """
      用循环替代递归，空间复杂度急剧降低，时间复杂度为o(n)
    """
    if n < 1:
      return 0
    if n == 1 or n == 2:
      return 1
    res = 1
    tmp1 = 0
    tmp2 = 1
    for _ in xrange(1, n):
      res = tmp1 + tmp2
      tmp1 = tmp2
      tmp2 = res
    return res

  def fibonacci3(self, n):
    """
      进一步减少迭代次数，采用矩阵求幂的方法，时间复杂度为o(log n)，当然了，这种方法需要额外计算矩阵，计算矩阵的时间开销没有算在内.其中还运用到了位运算。
    """
    base = [[1, 1], [1, 0]]
    if n < 1:
      return 0
    if n == 1 or n == 2:
      return 1
    res = self.__matrix_power(base, n-2)
    return res[0][0] + res[1][0]

  def __matrix_power(self, mat, n):
    """
      求一个方阵的幂
    """
    if len(mat) != len(mat[0]):
      raise ValueError("The length of m and n is different.")
    if n < 0 or str(type(n)) != "<type 'int'>":
      raise ValueError("The power is unsuitable.")
    product, tmp = [], []
    for _ in xrange(len(mat)):
      tmp.append(0)
    for _ in xrange(len(mat)):
      product.append(copy.deepcopy(tmp))
    for _ in xrange(len(mat)):
      product[_][_] = 1
    tmp = mat
    while n > 0:
      if (n & 1) != 0: # 按位与的操作，在幂数的二进制位为1时，乘到最终结果上，否则自乘
        product = self.__multiply_matrix(product, tmp)
      tmp = self.__multiply_matrix(tmp, tmp)
      n >>= 1
    return product

  @staticmethod
  def __multiply_matrix(mat1, mat2):
    """
      矩阵计算乘积
    :param m: 矩阵1，二维列表
    :param n: 矩阵2
    :return: 乘积
    """
    if len(mat1[0]) != len(mat2):
      raise ValueError("Can not compute the product of mat1 and mat2.")
    product, tmp = [], []
    for _ in xrange(len(mat2[0])):
      tmp.append(0)
    for _ in xrange(len(mat1)):
      product.append(copy.deepcopy(tmp))
    for i in xrange(0, len(mat1)):
      for j in xrange(0, len(mat2[0])):
        for k in xrange(0, len(mat1[0])):
          if mat1[i][k] != 0 and mat2[k][j] != 0:
            product[i][j] += mat1[i][k] * mat2[k][j]
    return product


f = Fibonacci()
print f.fibonacci1(23)
print f.fibonacci2(23)
mat1 = [[2,4,5],[1,0,2],[4,6,9]]
mat2 = [[2,9],[1,0],[5,7]]
print f.fibonacci3(23)