leetcode 368. Largest Divisible Subset 类似LISS最长递增子序列问题 + DP动态规划
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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]
仔细想一下,这道题可以转化为和LISS最长递增子序列一样的问题去使用DP动态规划求解,这道题很值得反思。
代码如下:
import java.util.ArrayList;import java.util.Arrays;import java.util.List;/* * 和LISS最长递增子序列做法思路完全一样 * * 1.先将数字排序; * 2.dp[i]表示nums[i]可以整除的最多数; * 3.有递推公式 dp[i]=max(dp[i],dp[j]+1) if dp[i]%dp[j]==0, j<i; * */class Solution { public List<Integer> largestDivisibleSubset(int[] nums) { List<Integer> res=new ArrayList<>(); if(nums==null || nums.length<=0) return res; if(nums.length==1) { res.add(nums[0]); return res; } int[] dp=new int[nums.length]; Arrays.fill(dp, 1); int [] pre=new int[nums.length]; for(int i=0;i<nums.length;i++) pre[i]=i; int maxLen=0,maxIndex=0; Arrays.sort(nums); for(int i=0;i<nums.length;i++) { for(int j=0;j<i;j++) { if(nums[i]%nums[j]==0 && dp[j]+1>dp[i]) { dp[i]=dp[j]+1; pre[i]=j; } } if(dp[i]>maxLen) { maxLen=dp[i]; maxIndex=i; } } int i=maxIndex; while(pre[i]!=i) { res.add(nums[i]); i=pre[i]; } res.add(nums[i]); return res; }}
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