254. 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:

Each combination's factors must be sorted ascending, for example: The factors of 2 and 6 is [2, 6], not [6, 2].
You may assume that n is always positive.
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]
]

链接： http://leetcode.com/problems/factor-combinations/

题解：

求一个数的所有factor，这里我们又想到了DFS + Backtracking，需要注意的是，factor都是>= 2的，并且在此题里，这个数本身不能算作factor，所以我们有了当n <= 1时的判断 if(list.size() > 1) add the result to res.

Time Complexity - O(2ⁿ)， Space Complexity - O(n).

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> list = new ArrayList<>();
        getFactors(res, list, n, 2);
        return res;
    }
    
    private void getFactors(List<List<Integer>> res, List<Integer> list, int n, int factor) {
        if(n <= 1) {
            if(list.size() > 1)
                res.add(new ArrayList<Integer>(list));
            return;
        }
        
        for(int i = factor; i <= n; i++) {
            if(n % i == 0) {
                list.add(i);
                getFactors(res, list, n / i, i);
                list.remove(list.size() - 1);
            }
        }
    }
}

二刷：

还是使用了一刷的办法，dfs + backtracking。但递归结束的条件更新成了n == 1。但是速度并不是很快，原因是没有做剪枝。我们其实可以设置一个upper limit，即当i > Math.sqrt(n)的时候，我们不能继续进行下一轮递归，此时就要跳出了。

Java:

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> res =  new ArrayList<>();
        if (n <= 1) return res;
        getFactors(res, new ArrayList<>(), n, 2);
        return res;
    }
    
    private void getFactors(List<List<Integer>> res, List<Integer> list, int n, int pos) {
        if (n == 1) {
            if (list.size() > 1) res.add(new ArrayList<>(list));
            return;
        }
        for (int i = pos; i <= n; i++) {
            if (n % i == 0) {
                list.add(i);
                getFactors(res, list, n / i, i);
                list.remove(list.size() - 1);
            }
        }
    }
}

Update: 使用@yuhangjiang的方法，只用计算 2到sqrt(n)的这么多因子，大大提高了速度。

public class Solution {
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> res =  new ArrayList<>();
        if (n <= 1) return res;
        getFactors(res, new ArrayList<>(), n, 2);
        return res;
    }
    
    private void getFactors(List<List<Integer>> res, List<Integer> list, int n, int pos) {
        for (int i = pos; i <= Math.sqrt(n); i++) {
            if (n % i == 0 && n / i >= i) {
                list.add(i);
                list.add(n / i);
                res.add(new ArrayList<>(list));
                list.remove(list.size() - 1);
                getFactors(res, list, n / i, i);
                list.remove(list.size() - 1);
            }
        }
    }
}

Reference:

https://leetcode.com/discuss/51261/iterative-and-recursive-python

https://leetcode.com/discuss/87926/java-2ms-easy-to-understand-short-and-sweet

https://leetcode.com/discuss/58828/a-simple-java-solution

https://leetcode.com/discuss/72224/my-short-java-solution-which-is-easy-to-understand

https://leetcode.com/discuss/82087/share-bit-the-thought-process-short-java-bottom-and-top-down