*Factor Combinations

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. Each combination's factors must be sorted ascending, for example: The factors of 2 and 6 is [2, 6], not [6, 2].
2. You may assume that n is always positive.
3. 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]
]

快速解法：

public class Solution {
public List<List<Integer>> getFactors(int n) {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    helper(result, new ArrayList<Integer>(), n, -1);
    return result;
}

public void helper(List<List<Integer>> result, List<Integer> item, int n, int start){
   if (start != -1) {
        item.add(n);
        result.add(new ArrayList<Integer>(item));
        item.remove(item.size() - 1);
    }
    
    for (int i = Math.max(start,2); i <= Math.sqrt(n); ++i) {
        if (n % i == 0) {
            item.add(i);
            helper(result, item, n/i, i);
            item.remove(item.size()-1);
        }
    }
}
}

请忽略一下解法

什么破方法慢死！recursion的解法：

public class Solution {
 public List<List<Integer>> getFactors(int n) {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    helper(result, new ArrayList<Integer>(), n, 2);
    return result;
}

public void helper(List<List<Integer>> result, List<Integer> item, int n, int start){
    if (n <= 1) {
        if (item.size() > 1) {
            result.add(new ArrayList<Integer>(item));
        System.out.println("add item to result");
        }
        return;
    }

    for (int i = start; i <= n; ++i) {
        if (n % i == 0) {
            item.add(i);
            System.out.println("item");
            System.out.println(item);
            helper(result, item, n/i, i);
            item.remove(item.size()-1);
            System.out.println("removed item");
            System.out.println(item);
        }
    }
}
}

运行结果：

test case n=8

item
[2]
item
[2, 2]
item
[2, 2, 2]
add to result
removed item
[2, 2]
removed item
[2]
item
[2, 4]
add to result
removed item
[2]
removed item
[]
item
[4]
removed item
[]
item
[8]
removed item
[]

结果： 

Your answer

[[2,2,2],[2,4]]


Expected answer

[[2,4],[2,2,2]]

reference：https://leetcode.com/discuss/51250/my-recursive-dfs-java-solution