LeetCode Factor Combinations

原题链接在这里：https://leetcode.com/problems/factor-combinations/

题目：

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]
]

题解：

DFS needs state current remaining number n, current starting factor, current item list and result.

For i from starting factor to n, check if n%i == 0. If yes, we could add i to current item list and dfs to next level.

starting factor could help maintain there is no duplicate.

e.g. n = 6. When it is divided by 3, the next factor needs to start from 3. Thus there is no [3,2] added since [2, 3] is already in the list. Could use such factor to maintain distinct.

Base case 是target变成了1, 并且item的size要大于1. 这里对item 的size 有要求是因为若初始target = 12, 返回结果是不应该包含{12}这个item的.

Time Complexity: Exponential.

Space: log(target).

AC Java:

 public class Solution {
     public List<List<Integer>> getFactors(int n) {
         List<List<Integer>> res = new ArrayList<List<Integer>>();
         if(n<=1){
             return res;
         }

         findFactors(n, 2, new ArrayList<Integer>(), res);
         return res;
     }

     //注意要写好的signature
     private void findFactors(int n, int factor, List<Integer> item, List<List<Integer>> res){
         //base case 不但要n==1, 还需要item.size() > 1, 否则n = 2, 就会添加一个item {2}.
         if(n == 1 && item.size() > 1){
             res.add(new ArrayList<Integer>(item));
             return;
         }
         for(int i = factor; i<=n; i++){
             if(n%i == 0){
                 item.add(i);
                 findFactors(n/i, i, item, res);
                 item.remove(item.size()-1);
             }
         }
     }
 }

类似Combination Sum.