LeetCode-Largest Divisble Subset

Given a set of distinct positive integers, find the largest subset such that every pair (S_i, S_j) of elements in this subset satisfies: S_i % S_j = 0 or S_j % S_i = 0.

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

Example 1:

nums: [1,2,3]

Result: [1,2] (of course, [1,3] will also be ok)

Example 2:

nums: [1,2,4,8]

Result: [1,2,4,8]

Credits:
Special thanks to @Stomach_ache for adding this problem and creating all test cases.

Analysis:

Sort the array, then for nums[i], if nums[i] % nums[j] == 0, then nums[i] is in the largest divisible set of nums[j]. We just need to find out the largest subset of nums[i].

Solution:

public class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        List<Integer> resList = new LinkedList<Integer>();        
        if (nums.length == 0) return resList;

        Arrays.sort(nums);
        int[] size = new int[nums.length];
        int[] pre = new int[nums.length];

        int maxSize = 0;
        int maxInd = -1;

        for (int i=0;i<nums.length;i++){
            int localMax = 0;
            int localInd = -1;
            for (int j=0;j<i;j++)
                if (nums[i] % nums[j] == 0 && localMax < size[j]+1){
                    localMax = size[j]+1;
                    localInd = j;
                }
            if (localInd == -1){
                localMax = 1;
                localInd = -1;
            } 
            size[i] = localMax;
            pre[i] = localInd;
            

            if (maxSize < localMax){
                maxSize = localMax;
                maxInd = i;
            }
        }
        
        resList.add(nums[maxInd]);
        int preInd = pre[maxInd];
        while (preInd != -1){
            maxInd = preInd;
            preInd = pre[maxInd];
            resList.add(nums[maxInd]);
        }
        Collections.reverse(resList);
        
        return resList;
    }
}