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:
- You may assume that n is always positive.
- Factors should be greater than 1 and less than n.
Example 1:
Input: 1
Output: []
Example 2:
Input: 37
Output:[]
Example 3:
Input: 12
Output:
[
[2, 6],
[2, 2, 3],
[3, 4]
]
Example 4:
Input:32
Output: [ [2, 16], [2, 2, 8], [2, 2, 2, 4], [2, 2, 2, 2, 2], [2, 4, 4], [4, 8] ]
class Solution { public List<List<Integer>> getFactors(int n) { List<List<Integer>> res = new ArrayList<>(); helper(res, new ArrayList<>(), n, 2); return res; } private void helper(List<List<Integer>> res, List<Integer> list, int n, int start) { if (n == 1) { // exclude the self case if (list.size() > 1) { res.add(new ArrayList<>(list)); return; } } for (int i = start; i <= n; i++) { if (n % i == 0) { list.add(i); helper(res, list, n / i, i); list.remove(list.size() - 1); } } } }
class Solution { public List<List<Integer>> getFactors(int n) { List<Integer> factors = new ArrayList<>(); for (int i = 2; i * i <= n; i++) { if (n % i == 0) { factors.add(i); if (i * i != n) { factors.add(n / i); } } } // Collections.sort(factors); // System.out.println(factors); List<List<Integer>> res = new ArrayList<>(); dfs(res, factors, new ArrayList<Integer>(), 1, n, 0); return res; } private void dfs(List<List<Integer>> res, List<Integer> factors, List<Integer> list, int cur, int n, int index) { if (cur > n) { return; } if (cur == n) { if (list.size() > 1) { res.add(new ArrayList<>(list)); } return; } // System.out.println(index); for (int i = index; i < factors.size(); i++) { list.add(factors.get(i)); dfs(res, factors, list, cur * factors.get(i), n, i); list.remove(list.size() - 1); } } }