LeetCode 368. Largest Divisible Subset

题目：Given a set of distinct positive integers, find the largest subset such that every pair (S_i, S_j) of elements in this subset satisfies: S_i % S_j = 0 or S_j % S_i = 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]

解题思路：使用动态规划解决问题。先对nums递增排序。首先，若nums[i]%nums[j]==0（i>j）,那么能整除nums[j]的都能整除nums[i]。我们用数组dp记录每个位置的最大可整除子集的元素个数，即dp[i]为集合{nums[0],...,nums[i]} 的最大可整除子集的元素个数，那么dp[i]=max(dp[j]+1,dp[i])(其中j为可整除nums[i]的元素的下标)，时间复杂度为O(N平方)，空间复杂度为O(N).

代码如下：

class Solution {public:    vector<int> largestDivisibleSubset(vector<int>& nums) {                int len = nums.size(),max,max_index,flag;        vector<int> dp(len),pos(len);        vector<int> ans;        if(nums.empty()) return ans;        sort(nums.begin(),nums.end());        for(int i = 0; i != len; ++i) pos[i]=i;        max=0,max_index=0;        for(int i = 0; i != len; ++i){            flag=0;            for(int j = 0; j < i; ++j){                if(!(nums[i]%nums[j])){                    if(dp[i]<dp[j]+1){                        dp[i] = dp[j]+1;                        pos[i] = j;                    }                    flag=1;                }            }            if(!flag){                dp[i]=1;                pos[i]=i;            }            if(dp[i]>max){                max = dp[i];                max_index=i;            }        }        int k = max_index;        while(pos[k]!=k){            ans.insert(ans.begin(),nums[k]);            k=pos[k];        }        ans.insert(ans.begin(),nums[k]);        return ans;    }};