Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,2]
, a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
这道题就是Subsets的延伸,只不过数组中有重复的数字。
所以每次遇到重复的数字,continue就好。(注意flag的设立,必须有flag,否则结果中会出现一个重复的数字都没有)。
public class Solution { List<List<Integer>> res = new ArrayList<>(); int[] num; int len; public List<List<Integer>> subsetsWithDup(int[] nums) { len = nums.length; num = nums; Arrays.sort(num); res.add(new ArrayList<>()); sub(0,new ArrayList<>() ); return res; } public void sub(int start,List<Integer> ans){ int flag = 0; for( int i = start;i<len;i++){ if( i > 0 && flag == 1 && num[i] == num[i-1]) continue; ans.add(num[i]); flag = 1; res.add(new ArrayList<Integer>(ans)); sub(i+1,ans); ans.remove(ans.size()-1); } return ; } }