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; } }