8.2—暴力枚举法—Subsets II

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本文介绍了一种算法，用于从可能包含重复元素的整数集合中找出所有可能的子集，并确保子集内部元素顺序非递减且不包含重复子集。通过具体的代码实现展示了如何使用 C++ 完成这一任务。

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描述
Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
Elements in a subset must be in non-descending order. e solution set must not contain duplicate
subsets. For example, If S = [1,2,2], a solution is:
[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
bool comp(const vector<int> &a, const vector<int> &b)
{
	return a.size() < b.size();
}
vector<vector<int>> Subsets(vector<int> data)
{
	vector<vector<int>> result;
	if (data.size() <= 0)
		return result;

	int length = data.size();
	int num = pow(2, length);
	int num1 = pow(2, length - 1);

	sort(data.begin(), data.end());
	for (int i = 0; i < num; i++)
	{
		vector<int> path;
		if (i == 0)
		{
			result.push_back(path);
			continue;
		}

		for (int j = 0; j < length; j++)
		{
			int temp = num1;
			temp = temp >>= j;
			if (i&temp)
			{
				path.push_back(data[j]);
			}
		}
		result.push_back(path);
	}
	sort(result.begin(), result.end(), comp);
	auto it = unique(result.begin(), result.end());
	result.resize(distance(result.begin(), it));
	return result;
}
int main()
{
	int a[3] = { 1, 2, 2 };
	vector<int> data(begin(a), end(a));
	vector<vector<int>> res = Subsets(data);
	for (int i = 0; i < res.size(); i++)
	{
		if (res[i].size() == 0)
			cout << "[]";

		for (int j = 0; j < res[i].size(); j++)
			cout << res[i][j] << " ";
		cout << endl;
	}
}