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leetcode dfs Subsets II

2024-07-26 17:59:48   219人阅读

Subsets II

 Total Accepted: 19243 Total Submissions: 71148My Submissions

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,
If S = [1,2,2], a solution is:

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

题意：给定一个可能包含重复整数的集合，返回集合的所有子集。要求返回的子集不可以重复。
思路：dfs
因为集合中的元素可能重复，所以相当于集合中的元素可以选择 0 次到若干次。
复杂度：时间O(n)，空间O(log n)

vector<vector<int> > res;

void dfs(const vector<int> &S, int cur, vector<int> subset){
	int size = S.size();
	if(cur == size){
		res.push_back(subset);
		return ;
	}
	int next = cur + 1;
	for(; next < size && S[next] == S[cur]; ++next);
	dfs(S, next, subset);
	for(int i = 0; i < next - cur; ++i){
		subset.push_back(S[cur]);
		dfs(S, next, subset);
	}
}

vector<vector<int> > subsetsWithDup(vector<int> &S) {
	sort(S.begin(), S.end());
	dfs(S, 0, vector<int>());
	return res;
}


leetcode dfs Subsets II


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