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],
  []
]

Solution：Backtracking

思路：
總結(jié)見：http://www.itdecent.cn/p/883fdda93a66
其實(shí)就是backtracking 的 simple組合問題 + 去重，套路ok，沒有終止條件，每一步都add到cur_res.
Time Complexity: O(2^n) Space Complexity: O(2N) 不算result, 含遞歸緩存和cur_res

Solution Code：

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

    private void backtrack(int [] nums, int start, List<Integer> cur_res, List<List<Integer>> result) {
        result.add(new ArrayList<>(cur_res));
        for(int i = start; i < nums.length; i++){
            if(i > start && nums[i] == nums[i - 1]) continue; // skip duplicates
            cur_res.add(nums[i]);
            backtrack(nums, i + 1, cur_res, result);
            cur_res.remove(cur_res.size() - 1);
        }
    } 
}