java dfs算法_java-如何找到dfs回溯算法的时间复杂度？

我正在尝试在leetcode https://leetcode.com/problems/factor-combinations/description/上解决此问题

Numbers can be regarded as product of its factors. For example

8 = 2 x 2 x 2; = 2 x 4.

编写一个接受整数n并返回其因子所有可能组合的函数.

虽然我可以使用dfs方法编写代码,但在输入方面,我仍然很难推动其最坏情况下的时间复杂性.谁能帮忙吗？

public List> getFactors(int n) {

List> result = new ArrayList>();

List current = new ArrayList();

getFactorsHelper(n,2,current,result);

return result;

}

public void getFactorsHelper(int n,int start,List current, List> result){

if(n<=1 && current.size()>1){

result.add(new ArrayList<>(current));

return;

}

for(int i=start;i<=n;i++){

if(n%i==0) {

current.add(i);

getFactorsHelper(n/i,i,current,result);

current.remove(current.size()-1);

}

}

}