Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
For example,
If S = [1,2,2]
, a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]递归+增量构造法
public class Solution { List<List<Integer>> res = new ArrayList<List<Integer>>(); List<Integer> list = new ArrayList<Integer>(); public List<List<Integer>> subsetsWithDup(int[] num) { Arrays.sort(num); help(num,0); return res; } private void help(int[]num,int start){ if(!res.contains(list)){ List<Integer> lin = new ArrayList<Integer>(list); res.add(lin); } for(int i=start;i<num.length;i++){ list.add(num[i]); help(num,i+1); list.remove(Integer.valueOf((num[i]))); } } }
原文地址:http://blog.csdn.net/guorudi/article/details/41253263