Leetcode90.Subsets_II

与Leetcode78.Subsets的思路完全相同，去重方法：将nums排序，当在某一次循环内遇到相同的数，则跳过。
时间复杂度: $O(Ne^N)$(此界随N增大而渐进)
C++代码：

class Solution {
public:
	vector<vector<int>> result;
	vector<vector<int>> subsetsWithDup(vector<int>& nums) {
		result.push_back({});
		sort(nums.begin(), nums.end());
		for (int i = 1; i <= nums.size(); i++)
			comb(nums.size(), i, {}, nums);
		return result;
	}
	void comb(int n, int k, vector<int> ans, vector<int>& nums)
	{
		if (k > n)
			return;
		if (k == 0)
		{
			result.push_back(ans);
			return;
		}
		if (k == n)
		{
			for (int i = k; i > 0; i--)
				ans.push_back(nums[i - 1]);
			result.push_back(ans);
			return;
		}
		for (int i = n; i >= k; i--)
		{
			if (i < n && nums[i - 1] == nums[i])
				continue;
			ans.push_back(nums[i - 1]);
			comb(i - 1, k - 1, ans, nums);
			ans.pop_back();
		}
	}
};