我写过这个算法,找到给定数字的所有因子并将它们放入一个列表中:
def find_all_factors(n):
factors = []
for i in range(1, floor(sqrt(n))+1):
if n % i == 0:
factors.append(i)
cofactor = n // i
if i != cofactor: factors.append(cofactor)
return factors
在列表中,cofactors将放在彼此旁边,但我希望它们以排序的顺序出现。上述算法的输出示例:对于n = 36
,它输出[1, 36, 2, 18, 3, 12, 4, 9, 6]
。我这样做是为了练习,我想知道按顺序排序的最有效方法是什么?有什么想法吗?
您可以在下面看到我的解决方案之一。它有效,但我认为它不是最佳的。
def find_all_factors(n):
lower_factors = []
higher_factors = []
for i in range(1, floor(sqrt(n))+1):
if n % i == 0:
lower_factors.append(i)
cofactor = n // i
if i != cofactor: higher_factors.append(cofactor)
return lower_factors + [higher_factors[-i] for i in range(1, len(higher_factors)+1)] #Reverses higher_factors.
答案 0 :(得分:1)
只需返回已排序的列表:
return sorted(factors)
如果您不喜欢使用sorted
函数,只需将循环范围更改为(1,n + 1):
def find_all_factors(n):
factors = []
for i in range(1, n+1):
if n % i == 0:
factors.append(i)
return factors
find_all_factors(12) # [1, 2, 3, 4, 6, 12]
另一种方法是使用bisect
(最有效的方法):
import bisect
def find_all_factors(n):
factors = []
for i in range(1, math.floor(math.sqrt(n))+1):
if n % i == 0:
bisect.insort(factors,i)
cofactor = n // i
if i != cofactor: bisect.insort(factors, cofactor)
return factors
find_all_factors(12) # [1, 2, 3, 4, 6, 12]
在此模块中插入O(n)
,但搜索为O(log(n))
答案 1 :(得分:1)
您缺少的一件事是在列表上操作更简单,更容易。有一个内置的Python用于反转序列:reversed
.
所以你可以这样做:
return lower_factors + list(reversed(higher_factors))
答案 2 :(得分:0)
from math import floor, sqrt
def find_all_factors(n):
factors = []
for i in xrange(1, int(floor(sqrt(n)))+1):
quotient, remainder = divmod(n, i)
if remainder == 0:
factors.append(i)
if quotient not in factors:
factors.append(quotient)
return sorted(factors)
print find_all_factors(36)
输出:[1, 2, 3, 4, 6, 9, 12, 18, 36]