【Leetcode】Largest Divisible Subset

题目链接：https://leetcode.com/problems/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]

思路：

dp

dp数组表示从0~i包括第i个元素最大的divisible subset size

pre数组用来标记 状态转移过程中的方向，用于回溯最大值时的解集。

dp[i]=max{dp[i],dp[j]+1}

算法：

public List<Integer> largestDivisibleSubset(int[] nums) {
LinkedList<Integer> res = new LinkedList<Integer>();
if (nums.length == 0) {
return res;
}
Arrays.sort(nums);

int dp[] = new int[nums.length]; //记录从0~i包括nums[i]的最大subset的size
int pre[] = new int[nums.length];//记录到当前元素最大size的前一位数的下标
int maxIdx = -1, max = -1;

for (int i = 0; i < nums.length; i++) { //初始化
dp[i] = 1;
pre[i] = -1;
}

for (int i = 1; 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;
pre[i] = j;//
}
}
}

for (int i = 0; i < nums.length; i++) {//找到最大的子集size 和它最后元素的下标
if (dp[i] > max) {
max = dp[i];
maxIdx = i;
}
}

for (int i = maxIdx; i >= 0;) { //回溯解集
res.addFirst(nums[i]);
i = pre[i];
}
return res;
}