[leedcode 90] Subsets II

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++;
        }
    }
}