Subsets II
问题：
Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
思路：
dfs + 回溯
我的代码：
public class Solution { public List<List<Integer>> subsetsWithDup(int[] num) { if(num == null || num.length == 0) return rst; List<Integer> list = new ArrayList<Integer>(); Arrays.sort(num); helper(list, num, 0); return rst; } private List<List<Integer>> rst = new ArrayList<List<Integer>>(); public void helper(List<Integer> list, int[] candidates, int start) { rst.add(new ArrayList(list)); for(int i = start ; i < candidates.length; i++) { if( i != start && candidates[i] == candidates[i-1]) continue; list.add(candidates[i]); helper(list, candidates, i + 1); list.remove(list.size() - 1); } } }
学习之处：
真正理解了程序，也就明白了if( i != start && candidates[i] == candidates[i-1]) continue;判断的精华所在了。
原文：http://www.cnblogs.com/sunshisonghit/p/4328430.html
