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: 

1. You may assume that n is always positive.
2. 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]]

  

算法：DFS。函数头：private void dfs(int target, int num, List<Integer> crt, List<List<Integer>> result) 

不断拆解为小一点的继续需要整除的数字。从num的数字开始，试着加入能被target整除的那些数字，来递归dfs(target / i, i, crt, result); 最后达到让target变成1的目标。

细节：用target % i == 0否来判断是否能够整除。

 

class Solution {    public List
      
       
        > getFactors(int n) {        List
        
         
          > result = new ArrayList<>();        if (n <= 1) {            return result;        }        dfs(n, 2, new ArrayList
          
           (), result);        return result;    }        private void dfs(int target, int num, List
           
             crt, List
            
             
              > result) { if (target == 1 && crt.size() > 1) { result.add(new ArrayList
              
               (crt)); return; } for (int i = num; i <= target; i++) { if (target % i != 0) { continue; } crt.add(i); dfs(target / i, i, crt, result); crt.remove(crt.size() - 1); } } }