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]
]
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：
class Solution {
public List<List<Integer>> getFactors(int n) {
List<List<Integer>> res = new ArrayList<>();
helper(res, new ArrayList<>(), n, 2);
return res;
}
public void helper(List<List<Integer>> res, List<Integer> list, int n, int start){
if(n == 1){
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);
}
}
}
}