LeetCode 368. Largest Divisible Subset

原题链接在这里：https://leetcode.com/problems/largest-divisible-subset/description/

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]

题解：

 求最长的subset, subset中每两个数大的除以小的余数是0. 储存到当前点, 包括当前点, 符合要求的最大subset长度.

Let count[i] denotes largest subset ending with i.

For all j < i, if nums[i] % nums[j] == 0, then i could be added after j and may construct a largeer subset. Update count[i]. 

But how to get the routine. Use preIndex[i] denotes the pre index of i in the largest subset. Then could start with last index and get pre num one by one. 

Time Complexity: O(n^2). n = nums.length.

Space: O(n).

AC Java:

class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        List<Integer> res = new ArrayList<Integer>();
        
        if(nums == null || nums.length == 0){
            return res;    
        }
        
        int len = nums.length;
        Arrays.sort(nums);
        int [] count = new int[len];
        int [] preIndex = new int[len];
        int resCount = 0;
        int resIndex = -1;
        
        for(int i = 0; i<len; i++){
            count[i] = 1;
            preIndex[i] = -1;
            for(int j = 0; j<i; j++){
               if(nums[i]%nums[j] == 0 && count[j]+1 > count[i]){
                   count[i] = count[j]+1;
                   preIndex[i] = j;
               } 
            }
            
            if(resCount < count[i]){
                resCount = count[i];
                resIndex = i;
            }
        }
        
        while(resIndex != -1){
            res.add(0, nums[resIndex]);
            resIndex = preIndex[resIndex];
        }
        
        return res;
    }
}

类似Longest Increasing Subsequence.