Factor Combinations

Factor Combinations

Problem:

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.
 1 public class Solution {
 2     public List<List<Integer>> getFactors(int n) {
 3         List<List<Integer>> listAll = new ArrayList<>();
 4         if (n < 2) return listAll;
 5         List<Integer> list = factors(n);
 6 
 7         helper(n, 1, 0, list, listAll, new ArrayList<Integer>());
 8         return listAll;
 9     }
10 
11     public void helper(int n, long value, int index, List<Integer> list, List<List<Integer>> listAll, List<Integer> temp) {
12         if (index >= list.size() || value >= n) return;
13 
14         int curValue = list.get(index);
15         temp.add(curValue);
16         value *= curValue;
17 
18         if (n == value) {
19             listAll.add(new ArrayList<Integer>(temp));
20         }
21 
22         helper(n, value, index, list, listAll, temp);
23         value /= curValue;
24         temp.remove(temp.size() - 1);
25         helper(n, value, index + 1, list, listAll, temp);
26     }
27 
28     public List<Integer> factors(int n) {
29         List<Integer> list = new ArrayList<>();
30         for (int i = 2; i <= (n + 1) / 2; i++) {
31             if (n % i == 0) {
32                 list.add(i);
33             }
34         }
35         return list;
36     }
37 }

上面代码其实可以用另下面这种方法，原理一样。

 1 public class Solution {
 2   public List<List<Integer>> getFactors(int n) {
 3       List<List<Integer>> result = new ArrayList<>();
 4       List<Integer> list = new ArrayList<>();
 5       helper(2, 1, n, result, list);
 6       return result;
 7   }
 8 
 9   public void helper(int start, int product, int n, List<List<Integer>> result, List<Integer> curr) {
10       if (start > n || product > n) return;
11 
12       if (product == n) {
13           List<Integer> t = new ArrayList<>(curr);
14           result.add(t);
15       }
16 
17       for (int i = start; i <= n / 2; i++) {
18           if(i * product > n) break;
19           if (n % i == 0) {
20               curr.add(i);
21               helper(i, i * product, n, result, curr);
22               curr.remove(curr.size() - 1);
23           }
24       }
25   }
26 }
posted @ 2016-08-05 03:25  北叶青藤  阅读(213)  评论(0)    收藏  举报