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]

一刷
題解：
先把數(shù)組排好序伍派。首先要明確，若a<b且b%a==0 , b <c 且 c%b==0那么必然有c%a==0
我們設(shè)dp[i] 為最大的子集長度诉植，更新的時候保存上一個的下標(biāo)即可。

public class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        List<Integer> res = new ArrayList<Integer>();
        if(nums.length == 0) return res;
        Arrays.sort(nums);
        int n = nums.length;
        int[] dp = new int[n], index = new int[n];
        Arrays.fill(dp, 1);
        Arrays.fill(index, -1);
        int max_index = 0, max_dp = 1;
        for(int i=0; i<n; i++){
            for(int j = i-1; j>=0; j--){
                if(nums[i] % nums[j] == 0 && dp[j]+1>dp[i]){
                    dp[i] = dp[j]+1;
                    index[i] = j;
                }
            }
            
            if(max_dp<dp[i]){
                max_dp = dp[i];
                max_index = i;
            }
        }
        
        for(int i= max_index; i!=-1; i = index[i]){
            res.add(nums[i]);
        }
        return res;
    }
}

二刷
同上舌稀，但是要記住用index array作為指針（鏈表）這個特點

class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        List<Integer> res = new ArrayList<>();
        int len = nums.length;
        if(len==0) return res;
        int[] dp = new int[len], index = new int[len];
        int max_len = 1, max_index = 0;
        Arrays.sort(nums);
        Arrays.fill(dp, 1);
        Arrays.fill(index, -1);
        for(int i=0; i<len; i++){
            for(int j=i-1; j>=0; j--){
                if(nums[i]%nums[j] == 0 && dp[i]<dp[j]+1){
                    dp[i] = dp[j]+1;
                    index[i] = j;
                }
            }
            if(dp[i]>max_len){
                max_len = dp[i];
                max_index = i;
            }
        }
        
        for(int i=max_index; i>=0; i=index[i]){
            res.add(nums[i]);
        }
        return res;
    }
}