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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。
1 public class Solution { 2 public List<Integer> largestDivisibleSubset(int[] nums) { 3 if (nums == null || nums.length == 0) return new ArrayList<Integer>(); 4 5 int n = nums.length; 6 Arrays.sort(nums); 7 int[] dp = new int[n]; 8 Arrays.fill(dp, 1); 9 int[] parent = new int[n]; 10 Arrays.fill(parent, -1);//当parent数组中某数为-1时，表示这个数自己是一个集合 11 int max = 1, max_index = 0; 12 for (int i = 1; i < n; i++) { //calculate dp[i] 13 for (int j = i - 1; j >= 0; j--) { //i > j 14 if (nums[i] % nums[j] == 0 && dp[i] < dp[j] + 1) { //positive distinct numbers in num 15 dp[i] = dp[j] + 1; 16 parent[i] = j; 17 if (dp[i] > max) { 18 max = dp[i]; 19 max_index = i; 20 } 21 } 22 } 23 } 24 return genResult(nums, parent, max_index); 25 } 26 27 public List<Integer> genResult(int[] nums, int[] parent, int max_index) { 28 List<Integer> result = new ArrayList<>(); 29 int iter = max_index; 30 while (iter != -1) { 31 result.add(nums[iter]); 32 iter = parent[iter]; 33 } 34 return result; 35 } 36 }
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5728292.html
