Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

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],
  []
]

思路：
将递归的输入为当前的结果集，当前的数组，和遍历的起始下标。在递归中将起始下标后的值依次加入当前结果集，并将结果集加入结果集数组中。如果遇到重复的数字，则继续遍历下一个数字，直至遍历结束。

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> fnl = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        helper(fnl, new ArrayList<Integer>(), nums, 0);
        return fnl;
    }

    public void helper(List<List<Integer>> fnl, List<Integer> temp, int[] nums, int start){
        fnl.add(new ArrayList<Integer>(temp));
        for(int i = start; i < nums.length; i++){
            if(i > start && nums[i] == nums[i-1]) 
                continue;
            temp.add(nums[i]);
            helper(fnl, temp, nums, i+1);
            temp.remove(temp.size()-1);
        }
    }
}