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.
Analysis:
Sort the array, then for nums[i], if nums[i] % nums[j] == 0, then nums[i] is in the largest divisible set of nums[j]. We just need to find out the largest subset of nums[i].
Solution:
1 public class Solution { 2 public List<Integer> largestDivisibleSubset(int[] nums) { 3 List<Integer> resList = new LinkedList<Integer>(); 4 if (nums.length == 0) return resList; 5 6 Arrays.sort(nums); 7 int[] size = new int[nums.length]; 8 int[] pre = new int[nums.length]; 9 10 int maxSize = 0; 11 int maxInd = -1; 12 13 for (int i=0;i<nums.length;i++){ 14 int localMax = 0; 15 int localInd = -1; 16 for (int j=0;j<i;j++) 17 if (nums[i] % nums[j] == 0 && localMax < size[j]+1){ 18 localMax = size[j]+1; 19 localInd = j; 20 } 21 if (localInd == -1){ 22 localMax = 1; 23 localInd = -1; 24 } 25 size[i] = localMax; 26 pre[i] = localInd; 27 28 29 if (maxSize < localMax){ 30 maxSize = localMax; 31 maxInd = i; 32 } 33 } 34 35 resList.add(nums[maxInd]); 36 int preInd = pre[maxInd]; 37 while (preInd != -1){ 38 maxInd = preInd; 39 preInd = pre[maxInd]; 40 resList.add(nums[maxInd]); 41 } 42 Collections.reverse(resList); 43 44 return resList; 45 } 46 }
时间: 2024-11-29 07:20:13