leetcode JAVA Subsets II 4.26 难度系数4

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],
[]
]

public class Solution {public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {ArrayList<ArrayList<Integer>> results  = new ArrayList<ArrayList<Integer>>();ArrayList<Integer> result = new ArrayList<Integer>();Arrays.sort(num);dfs(num,0,result,results);return results;}private void dfs(int[] num, int step, ArrayList<Integer> result,ArrayList<ArrayList<Integer>> results) {if(!results.contains(result))results.add(new ArrayList<Integer>(result));for(int i=step;i<num.length;i++){result.add(num[i]);dfs(num, i+1, result, results);result.remove(result.size()-1);}}}

跟上题一样，可以2种方式加上去除重复。

第二种去重：

public class Solution {public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {ArrayList<ArrayList<Integer>> results  = new ArrayList<ArrayList<Integer>>();ArrayList<Integer> result = new ArrayList<Integer>();Arrays.sort(num);dfs(num,0,result,results);return results;}private void dfs(int[] num, int step, ArrayList<Integer> result,ArrayList<ArrayList<Integer>> results) {results.add(new ArrayList<Integer>(result));for(int i=step;i<num.length;i++){if(i!=step&&num[i]==num[i-1])continue;result.add(num[i]);dfs(num, i+1, result, results);result.remove(result.size()-1);}}}