LeetCode No.368. 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]

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题目链接：https://leetcode.com/problems/largest-divisible-subset/

题目大意：求出数组中最长非严格递增子序列，使得序列中任意相邻一对数值满足Si % Sj = 0 或者 Sj % Si = 0。

思路：动态规划，dp[i]记录nums[i]作为序列最后一位时的满足条件的最长子序列长度。

参考代码：

class Solution {public:    vector<int> largestDivisibleSubset(vector<int>& nums) {        int n = nums.size() ;        vector <int> ans ;        if ( n == 0 )            return ans ;        sort ( nums.begin() , nums.end() ) ;//从小到大排序        vector <int> pre ( n , 0 ) ;//记录序列前一个的下标        vector <int> dp ( n , 0 ) ;//记录nums[i]个为序列最后一位时的序列长度        pre[0] = 0 ;        dp[0] = 1 ;        for ( int i = 1 ; i < n ; i ++ )        {            int p = i , sum = 1 ;            for ( int j = 0 ; j < i ; j ++ )//寻找最长序列并记录前一个下标            {                if ( nums[i] % nums[j] == 0 || nums[j] % nums[i] == 0 )                {                    if ( dp[j] + 1 > sum )                    {                        sum = dp[j] + 1 ;                        p = j ;                    }                }            }            dp[i] = sum ;            pre[i] = p ;        }        int as = 0 , temp = 0 ;        for ( int i = 0 ; i < n ; i ++ )//找出最长序列最后一位下标        {            if ( dp[i] > as )            {                as = dp[i] ;                temp = i ;            }        }        while ( pre[temp] != temp ) //从后往前推，直到pre[temp] == temp说明到达序列第一位        {            ans.push_back ( nums[temp] ) ;            temp = pre[temp] ;        }         ans.push_back ( nums[temp] ) ;        reverse ( ans.begin() , ans.end() ) ;//结果逆序        return ans ;    }};