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]

分析：

思路: 将数组排序，然后用dp[i]表示从0到i最大的集合。为了得到dp[i]的值, 我们从i - 1 到 0 看是否 nums[i] % nums[j] ==0,  如果是，dp[i] = max(dp[i], dp[j]+1), 因为数组按照降序排序, 所以nums[j] < nums[i],并且之前能够被nums[j]整除的数, 也必然能够被 nums[i]整除。

 public class Solution {
     public List<Integer> largestDivisibleSubset(int[] nums) {
         if (nums == null || nums.length == ) return new ArrayList<Integer>();

         int n = nums.length;
         Arrays.sort(nums);
         int[] dp = new int[n];
         Arrays.fill(dp, );
         int[] parent = new int[n];
         Arrays.fill(parent, -);//当parent数组中某数为-1时，表示这个数自己是一个集合
         int max = , max_index = ;
         for (int i = ; i < n; i++) {   //calculate dp[i]
             for (int j = i - ; j >= ; j--) {  //i > j
                 if (nums[i] % nums[j] ==  && dp[i] < dp[j] + ) {    //positive distinct numbers in num
                     dp[i] = dp[j] + ;
                     parent[i] = j;
                     if (dp[i] > max) {
                         max = dp[i];
                         max_index = i;
                     }
                 }
             }
         }
         return genResult(nums, parent, max_index);
     }

     public List<Integer> genResult(int[] nums, int[] parent, int max_index) {
         List<Integer> result = new ArrayList<>();
         int iter = max_index;
         while (iter != -) {
             result.add(nums[iter]);
             iter = parent[iter];
         }
         return result;
     }
 }