leetcode 368 divided by the maximum number of sub-sets (dynamic programming)

Divisible by each other? Maximum number!

Title :
given a set of positive integers without duplication composition, divisible find the largest subset, subset any pair (Si, Sj) must be satisfied: Si% Sj = 0 or Sj% Si = 0.

Sample :
Example 1:
Input: [1,2,3]
Output: [1,2] (Of course, [1,3] also correct)
Example 2:
Input: [1,2,4,8]
Output: [ 1,2,4,8]

Topic analysis

Suppose the current maximum number divisible by n subset
if the subset can occur x == All numbers divisible> n + 1
appeared! State transition - so use dp Solution

Process Analysis

Set DP [i] represents the maximum number of position i divisible subset
max marking the maximum number of the subset
max_id number of marker subsets maximum starting
parent [i] on a position of the mark is divisible i

• First of nums from small to large
  (nums [i] <nums [ j] ==> nums [j] be nums [i] is divisible, then it is divisible by nums [i] is able divisible nums [j])
• traversing from 0 i
  j i from traversing
  if the nums [j] be the nums [i] divisible ==> DP [j] DP = [i] +. 1
  parent [i] = j - recording the image of position i a divisible j
• If mx <dp [j]
  to update and mx mx_id
• Finally, according to the answer mx join

code show as below:

class Solution {
public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<int> dp(nums.size(), 0), parent(nums.size(), 0), res;
        int mx = 0, mx_idx = 0;
        for (int i = nums.size() - 1; i >= 0; --i) {
            for (int j = i; j < nums.size(); ++j) {
                if (nums[j] % nums[i] == 0 && dp[i] < dp[j] + 1) {
                    dp[i] = dp[j] + 1;
                    parent[i] = j;
                    if (mx < dp[i]) {
                        mx = dp[i];
                        mx_idx = i;
                    }
                }
            }
        }
        for (int i = 0; i < mx; ++i) {
            res.push_back(nums[mx_idx]);
            mx_idx = parent[mx_idx];
        }
        return res;
    }
};