标签:des style blog color os io java ar for
Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example, If S = [1,2,2], a solution is: [ [2], [1], [1,2,2], [2,2], [1,2], [] ]
与Subsets问题的唯一区别在于,一个满足条件的set加入到最终结果sets里面的时候,需要先检查sets里面是否已经存在这个set,如果重复,则不添加。
这还是一道列举所有case的NP的题,用recursion, 用arraylist
1 import java.util.*; 2 3 public class Solution { 4 public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) { 5 java.util.Arrays.sort(num); 6 ArrayList<Integer> set = new ArrayList<Integer>(); 7 ArrayList<ArrayList<Integer>> sets = new ArrayList<ArrayList<Integer>>(); 8 9 for (int i = 0; i <= num.length; i++) { 10 helper(set, sets, num, i, 0); 11 } 12 return sets; 13 } 14 15 public void helper(ArrayList<Integer> set, ArrayList<ArrayList<Integer>> sets, int[] num, int i, int start) { 16 if (set.size() == i) { 17 if (!sets.contains(set)) { 18 sets.add(new ArrayList<Integer>(set)); 19 } 20 return; 21 } 22 for (int k = start; k < num.length; k++) { 23 set.add(num[k]); 24 helper(set, sets, num, i, k+1); 25 set.remove(set.size() - 1); 26 } 27 } 28 }
标签:des style blog color os io java ar for
原文地址:http://www.cnblogs.com/EdwardLiu/p/3958971.html