Leetcode: 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:

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]

DP Solution similar to Longest Increasing Subsequence:

我的解法：用一个arraylist存以某一个element结尾的最长sequence

public class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        List<Integer> res = new ArrayList<Integer>();
        if (nums==null || nums.length==0) return res;
        int[] dp = new int[nums.length];
        int maxdpIndex = 0;
        ArrayList<ArrayList<Integer>> records = new ArrayList<ArrayList<Integer>>();
        
        Arrays.sort(nums);
        for (int i=0; i<nums.length; i++) {
            dp[i] = 1;
            ArrayList<Integer> record = new ArrayList<Integer>();
            for (int j=0; j<i; j++) {
                if (nums[i] % nums[j] == 0) {
                    if(dp[j] + 1 > dp[i]) {
                        dp[i] = dp[j] + 1;
                        record.clear();
                        record.addAll(records.get(j));
                    }
                }
            }
            record.add(nums[i]);
            records.add(new ArrayList<Integer>(record));
            if (dp[i] > dp[maxdpIndex]) maxdpIndex = i;
        }
        res = records.get(maxdpIndex);
        return res;
    }
}

别人的好方法，大体思路一样，只是不存arraylist, 而用一个preIndex数组存以某一个element结尾的最长sequence的上一个元素index, 这样做快了很多

public class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        int n = nums.length;
        int[] count = new int[n];
        int[] pre = new int[n];
        Arrays.sort(nums);
        int max = 0, index = -1;
        for (int i = 0; i < n; i++) {
            count[i] = 1;
            pre[i] = -1;
            for (int j = i - 1; j >= 0; j--) {
                if (nums[i] % nums[j] == 0) {
                    if (1 + count[j] > count[i]) {
                        count[i] = count[j] + 1;
                        pre[i] = j;
                    }
                }
            }
            if (count[i] > max) {
                max = count[i];
                index = i;
            }
        }
        List<Integer> res = new ArrayList<>();
        while (index != -1) {
            res.add(nums[index]);
            index = pre[index];
        }
        return res;
    }
}