# 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:**

```
Input: 
[1,2,3]
Output: 
[1,2] 
(of course, [1,3] will also be ok)
```

**Example 2:**

```
Input: 
[1,2,4,8]
Output: 
[1,2,4,8]
```

分析

区间型dp

本题结果无需连续subset即可 无需subarray

因为需要返回List,这里dp count 以i结尾的list, **index**和**pre**记载前数的数组和最后尾index,用来打印结果

一个count count\[i]\<count\[j]+1就可以更新。每次只需要比较旧数组里最后元素j是否能被新加入i %，不需要遍历旧数组所有元素

```
class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        n = len(nums)
        count = [0]*(n)
        pre = [-1]*(n)
        maxlen,index = float('-inf'),-1
        nums.sort()#要排序
        for i in range(n):
            for j in range(i-1,-1,-1):
                if nums[i]%nums[j]==0 and count[j]+1>count[i]:#只要能%最后元素j就行，不必所有元素都%.
                    count[i] = count[j]+1
                    pre[i] = j
            if count[i] > maxlen:
                maxlen = count[i]
                index = i
        res = []
        while index!=-1:
            res.append(nums[index])
            index = pre[index]            
        return res
```

用set慢慢扩展

```
class Solution:
    def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
        n = len(nums)
        count = [0]*(n)
        pre = [-1]*(n)
        maxlen,index = float('-inf'),-1
        nums.sort()#要排序
        s = {-1:set()}
        for x in nums:
            s[x] = max((s[d] for d in s if x%d == 0),key = len)|{x}
        return list(max(s.values(),key = len))
```
