?LeetCode刷題實(shí)戰(zhàn)368：最大整除子集數(shù)
Given a set of distinct positive integers nums, return the largest subset answer such that every pair (answer[i], answer[j]) of elements in this subset satisfies:
answer[i] % answer[j] == 0, or
answer[j] % answer[i] == 0
If there are multiple solutions, return any of them.
示例
示例 1：
輸入：nums = [1,2,3]
輸出：[1,2]
解釋：[1,3] 也會(huì)被視為正確答案。
示例 2：
輸入：nums = [1,2,4,8]
輸出：[1,2,4,8]
解題
class Solution {
public List<Integer> largestDivisibleSubset(int[] nums) {
int len = nums.length;
Arrays.sort(nums);
// 第 1 步：動(dòng)態(tài)規(guī)劃找出最大子集的個(gè)數(shù)、最大子集中的最大整數(shù)
int[] dp = new int[len];
Arrays.fill(dp, 1);
int maxSize = 1;
int maxVal = dp[0];
for (int i = 1; i < len; i++) {
for (int j = 0; j < i; j++) {
// 題目中說「沒有重復(fù)元素」很重要
if (nums[i] % nums[j] == 0) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
if (dp[i] > maxSize) {
maxSize = dp[i];
maxVal = nums[i];
}
}
// 第 2 步：倒推獲得最大子集
List<Integer> res = new ArrayList<Integer>();
if (maxSize == 1) {
res.add(nums[0]);
return res;
}
for (int i = len - 1; i >= 0 && maxSize > 0; i--) {
if (dp[i] == maxSize && maxVal % nums[i] == 0) {
res.add(nums[i]);
maxVal = nums[i];
maxSize--;
}
}
return res;
}
}