程序媛詹妮弗
终身学习

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.

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]
]

 

题意:

给定正整数n,返回所有乘积等于n的不同组合(除了n = n)。为避免重复,所有乘数单调递增。

 

思路:

backtracking

 

代码:

 1 class Solution {
 2     public List<List<Integer>> getFactors(int n) {
 3         List<List<Integer>> res = new ArrayList<>();
 4         helper(res, new ArrayList<>(), n, 2);
 5         return res;
 6     }
 7     
 8     public void helper(List<List<Integer>> res, List<Integer> list, int n, int start){
 9         if(n == 1){
10             if(list.size() > 1){
11                 res.add(new ArrayList<>(list));
12                 return;
13             }
14         }
15         for(int i = start; i< = n; i++){
16             if(n % i== 0){
17                 list.add(i);
18                 helper(res, list, n/i, i);
19                 list.remove(list.size()-1);
20             }
21         }
22     }
23 }

 

posted on 2018-06-09 05:08  程序媛詹妮弗  阅读(182)  评论(0编辑  收藏  举报