Leetcode - Factor Combination

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.

[分析]
依次判断2到n的平方根是否能被n整除，如果可以整除则当前 i 和 n / i是一个可行解，然后递归获取 n / i的所有因子组合，它们将与 i 一起组合成 n 的因子组合。
为避免得到重复解并满足因子组合各项从小到大排列，有两个注意点：
1）如果 i * i > n / i，无需递归，因为 n / i分解所得因子必然小于i，不符合要求。
2）递归 n / i时最小因子要从 i开始

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        return getFactorsWithStartParam(n, 2);
    }
    public List<List<Integer>> getFactorsWithStartParam(int n, int start) {
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        if (n < 4) return ret;
        int sqrt = (int)Math.sqrt(n);
        for (int i = start; i <= sqrt; i++) {
            if (n % i == 0) {
                int factor2 = n / i;
                List<Integer> item = new ArrayList<Integer>();
                item.add(i);
                item.add(factor2);
                ret.add(item);
                if (i * i <= factor2) {// avoid get smaller factor than i
                    for (List<Integer> subitem : getFactorsWithStartParam(factor2, i)) {// avoid get smaller factor than i
                        subitem.add(0, i);
                        ret.add(subitem);
                    }
                }
            }
        }
        return ret;
    }
    
}