DP-368-Largest Divisible Subset

Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies:

Si % Sj = 0 or Sj % Si = 0.

If there are multiple solutions, return any subset is fine.

Example 1:

Input: [1,2,3]
Output: [1,2] (of course, [1,3] will also be ok)
Example 2:

Input: [1,2,4,8]
Output: [1,2,4,8]
=============================================================
class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        if(nums.length == 0)
            return new LinkedList<Integer>();
        LinkedList<LinkedList<Integer>> res = new LinkedList<>();
        Arrays.sort(nums);
        
        LinkedList<Integer> set = new LinkedList<>();
        set.addFirst(Integer.valueOf(nums[nums.length - 1]));
        res.addFirst(set);
        for(int i = nums.length - 2; i > -1; i--){
            boolean found = false;
            int maxSize = -1;
            LinkedList<Integer> maxSet = null;
            for(Iterator<LinkedList<Integer>> iter = res.iterator(); iter.hasNext(); ){
                set = iter.next();
                if(set.getFirst() % nums[i] == 0 && set.size() > maxSize){
                    found = true;
                    maxSize = set.size();
                    maxSet = set;
                }
            }
            if(!found){
                set = new LinkedList<>();
                set.addFirst(Integer.valueOf(nums[i]));
                res.addFirst(set);
            } else {
                set = (LinkedList<Integer>) maxSet.clone();
                res.addFirst(set);
                set.addFirst(Integer.valueOf(nums[i]));
            }
            
        }    
        
        int maxSize = -1;
        List<Integer> finalRes = null;
        for(Iterator<LinkedList<Integer>> iter = res.iterator(); iter.hasNext(); ){
            List<Integer> l = iter.next();
            if(l.size() > maxSize){
                maxSize = l.size();
                finalRes = l;
            }
        }
        return finalRes;
    }
}

//6%