# Question

Formatted question description: https://leetcode.ca/all/254.html

```
254 Factor Combinations
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.
Note:
Each combination's factors must be sorted ascending, for example: The factors of 2 and 6 is [2, 6], not [6, 2].
You may assume that n is always positive.
Factors should be greater than 1 and less than n.
Examples:
input: 1
output:
[]
input: 37
output:
[]
input: 12
output:
[
[2, 6],
[2, 2, 3],
[3, 4]
]
input: 32
output:
[
[2, 16],
[2, 2, 8],
[2, 2, 2, 4],
[2, 2, 2, 2, 2],
[2, 4, 4],
[4, 8]
]
```

# Algorithm

Iterate from 2 to n. If the current number i is divisible by n, it means that i is a factor of n. Store it in a `onePath`

list, and then recursively call n/i. At this time, do not traverse from 2. It traverses from i to n/i, and the condition for stopping is when n is equal to 1, if there is a factor in `onePath`

at this time, store this combination in the result.

### Pitfall

Need to avoid case: n=32, and only 32 in this list, add check
`if (onePath.size() > 1)`

# Code

Java