368. Largest Divisible Subset-优快云博客

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/67925130

本文介绍了一种使用动态规划解决寻找最大整除子集问题的方法。通过将输入数组进行排序，并利用count数组记录可能的最大子集长度，最终找出满足条件的最大子集。附带示例代码帮助理解。

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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]

Credits:
Special thanks to @Stomach_ache for adding this problem and creating all test cases.

这道题用dp来做。先把array升序排列，新建两个数组，count[]存nums每个位置最多的余数，pre[]存之前能整出的nums的index。对于每一个nums的数，如果能整除nums，而且之前的count也比现在大，就替换掉现在的count。代码如下：

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