forked from keon/algorithms
-
Notifications
You must be signed in to change notification settings - Fork 2
/
all_factors.py
73 lines (59 loc) · 1.37 KB
/
all_factors.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
"""
Numbers can be regarded as product of its factors. For example,
8 = 2 x 2 x 2;
= 2 x 4.
Write a function that takes an integer n and return all possible combinations
of its factors.Numbers can be regarded as product of its factors. For example,
8 = 2 x 2 x 2;
= 2 x 4.
Examples:
input: 1
output:
[]
input: 37
output:
[]
input: 32
output:
[
[2, 16],
[2, 2, 8],
[2, 2, 2, 4],
[2, 2, 2, 2, 2],
"""
def get_factors(n):
def factor(n, i, combi, res):
while i * i <= n:
if n % i == 0:
res += combi + [i, int(n/i)],
factor(n/i, i, combi+[i], res)
i += 1
return res
return factor(n, 2, [], [])
def get_factors_iterative1(self, n):
todo, res = [(n, 2, [])], []
while todo:
n, i, combi = todo.pop()
while i * i <= n:
if n % i == 0:
res += combi + [i, n/i],
todo += (n/i, i, combi+[i]),
i += 1
return res
def get_factors_iterative2(n):
ans, stack, x = [], [], 2
while True:
if x > n / x:
if not stack:
return ans
ans.append(stack + [n])
x = stack.pop()
n *= x
x += 1
elif n % x == 0:
stack.append(x)
n /= x
else:
x += 1
if __name__ == "__main__":
print(get_factors(32))