动态规划-368. Largest Divisible Subset

题目：

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]

题意解读：给定一个含有不同正整数的集合，寻找最大的子集，该子集的特点是对于子集中任意两个数，都可以被整除。如果有多个子集存在，返回一种就可以

class Solution {    public List<Integer> largestDivisibleSubset(int[] nums) {        List<Integer> res = new ArrayList<Integer>();        if(nums.length == 1){            res.add(nums[0]);            return res;        }        if(nums.length == 0){            return res;        }                    //先对数组排序        Arrays.sort(nums);        //定义辅助数组        //dp[i]:到第i个位置LargestDivisibleSubset的长度        //状态转移方程：if(nums[i] % nums[j] == 0 && dp[i] < dp[j]+1) dp[i] = dp[j]+1        //pre[i]指向第i个元素的被除数的索引        int[] dp = new int[nums.length];int[] pre = new int[nums.length];        Arrays.fill(dp,1); Arrays.fill(pre,-1);        int max = 0; int maxid = 0;//首先将maxid随意指向一个元素        for(int i = 0; i < nums.length; i++){            for(int j = 0; j < i; j++){                if(nums[i] % nums[j] == 0 && dp[i] < dp[j]+1){                    dp[i] = dp[j] + 1;//更新dp[i]                    pre[i] = j;       //更新pre[i]                    if(dp[i] > max){                        max = dp[i];//更新最大长度                        maxid = i;  //更新最大长度对应的索引                    }                }            }        }        while(maxid != -1){            res.add(nums[maxid]);            maxid = pre[maxid];        }        return res;    }}