[leedcode 90] Subsets II

Given a collection of integers that might contain duplicates, nums, 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 nums =

[1,2,2]

, a solution is:

[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]

public class Solution {
List<Integer> seq;
List<List<Integer>> res;
public List<List<Integer>> subsetsWithDup(int[] nums) {
//本题和Subsets的不同就是需要对重复值进行处理，增加了一个判断，
//需要注意判断的范围，i首先要满足i>start
seq=new ArrayList<Integer>();
res=new ArrayList<List<Integer>>();
Arrays.sort(nums);
getSub(nums,0,nums.length);
return res;
}
public void getSub(int[] nums,int start,int end){
if(start<=end){
res.add(new ArrayList<Integer>(seq));
}
for(int i=start;i<end;i++){
if(i>start&&nums[i]==nums[i-1])continue;
seq.add(nums[i]);
getSub(nums,i+1,end);
seq.remove(seq.size()-1);
//while(i<nums.length-1&&nums[i]==nums[i+1]) i++;
}
}
}