Largest Divisible Subset

题目链接: https://leetcode.com/problems/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.

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]

思路: 可以用动态规划来解决. 状态转移方程也很容易得到, 也就是当前数组如果能够整除以前的数字, 那么就可以用这个状态方程: dp[i] = max(dp[i], dp[j]+1);

然后还要知道最长的集合, 那么只需要记录一下第一个到达最大长度的那个数字位置即可, 然后找到这个位置之后再遍历一遍数组, 只要能够整除这个数的就加入结果集合中.

另外为了保证数组按照升序排列, 而且他也没提有序, 那就先排个序吧.

代码如下:

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