Write a function that takes an integer n and return all possible combinations of its factors.

  1. You may assume that n is always positive.
  2. Factors should be greater than 1 and less than n.
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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]
]

其实从上面看出，1, n都不能当作factors, 所以arr.size()至少是2: i.e. 递归终止的条件

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class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> list = new ArrayList<>();
        dfs(list, new ArrayList<>(), n, 2); // 1 < x < n
        return list;
    }
    
    public void dfs(List<List<Integer>> list, List<Integer> arr, int n, int start) {
        if (n == 1 && arr.size() > 1) list.add(new ArrayList<>(arr));
        for (int i = start; i <= n; ++i) {
            if (n % i == 0) {
                arr.add(i);
                dfs(list, arr, n / i, i);
                arr.remove(arr.size() - 1);
            }
        }
    }
}

也可以看出, factor combination实际上还是递归就完了，只不过-变成了/, 并没有必要先全排列所有的因子。