LeetCode-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. You may assume that n is always positive.
2. 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]
]

Analysis:

Recursively build  a combination, for a target value (named target), we starting from the smallest factor, which is the last factor used rather than 2. If we get a new factor, saying target = x*y, we should guarantee that x<=y; we then go to the next layer of recursion looking for all combinations with target value of y.

 

Solution:

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        Deque<Integer> combination = new LinkedList<Integer>();
        List<List<Integer>> resList = new LinkedList<List<Integer>>();
        getFactorsRecur(n,n,2,combination,resList);
        return resList;
    }
    
    public void getFactorsRecur(int n, int target, int lastFactor, Deque<Integer> combination, List<List<Integer>> resList){
        if (target<lastFactor){
            return;
        }
        
        // add current combination into resList.
        if (target<n){
            List<Integer> temp = new ArrayList<Integer>();
            temp.addAll(combination);
            temp.add(target);
            resList.add(temp);
        }
        
        // Further decompose the current target.
        for (int i=lastFactor;i*i<=target;i++){
            if (target%i==0){
                combination.addLast(i);
                getFactorsRecur(n,target/i,i,combination,resList);
                combination.removeLast();
            }
        }
    }
}