Leetcode Subsets II 有重复元素的组合-优快云博客

本文介绍了一种解决含重复元素集合的所有子集问题的算法。通过先排序输入数组，然后利用unordered_set容器来确保结果中没有重复的子集。文章提供了两种实现方式，并详细展示了如何在遍历过程中插入新子集的同时避免重复。

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Subsets II

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.
The solution set must not contain duplicate subsets.

For example,
IfS=[1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

禁止重复，就使用set，map等容器，就可以很简单解决了。

class Solution {
public:
	vector<vector<int> > subsetsWithDup(vector<int> &S) 
	{
		sort(S.begin(), S.end());
		vector<vector<int> > rs(1);
		vector<int> temp;
		set<vector<int> > us;
		us.insert(temp);

		for (int i = 0; i < S.size(); i++)
		{
			for (int j = rs.size()-1; j>=0; j--)
			{
				temp = rs[j];
				temp.push_back(S[i]);
				rs.push_back(temp);
				us.insert(temp);
			}
		}
		rs.clear();
		rs.assign(us.begin(), us.end());
		return rs;
	}

	vector<vector<int> > subsetsWithDup2(vector<int> &S) 
	{
		sort(S.begin(), S.end());
		vector<vector<int> > rs(1);
		vector<int> temp;
		set<vector<int> > us;
		us.insert(temp);

		for (int i = 0; i < S.size(); i++)
		{
			for (int j = rs.size()-1; j>=0; j--)
			{
				temp = rs[j];
				temp.push_back(S[i]);
				rs.push_back(temp);
				us.insert(temp);
			}
			rs.assign(us.begin(),us.end());
		}
		return rs;
	}
};

//2014-2-14 update
	vector<vector<int> > subsetsWithDup(vector<int> &S) 
	{
		sort(S.begin(), S.end());
		vector<vector<int> > rs(1);
		unordered_set<vector<int> > sv(rs.begin(), rs.end());
		for (int i = 0; i < S.size(); i++)
		{
			for (int j = rs.size() - 1; j >= 0 ; j--)
			{
				rs.push_back(rs[j]);
				rs.back().push_back(S[i]);
				sv.insert(rs.back());
			}
			rs.assign(sv.begin(), sv.end());
		}
		return rs;
	}