leetcode 90

最新推荐文章于 2023-12-01 12:28:16 发布

原创最新推荐文章于 2023-12-01 12:28:16 发布 · 204 阅读

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与subsets类似，但是需要跳过重复的节点，加入while(level+1<nums.size()&&nums[level]==nums[level+1])level++;

                        []        
                   /          \        
                  /            \     
                 /              \
              [1]                []
           /       \           /    \
          /         \         /      \        
       [1 2]       [1]       [2]     []
      /     \     /   \     /   \    / \
  [1 2 2] [1 2]  X   [1]  [2 2] [2] X  []

如图所示，只需跳过未添加重复节点的节点如[1],[]

void DFS(vector<vector<int>>& res, vector<int>& out, vector<int>& nums,int level) {
	if (level == nums.size()) {
		res.push_back(out);
		return;
	}
	else
	{
		out.push_back(nums[level]);
		DFS(res, out, nums, level+1);
		out.pop_back();
        while(level+1<nums.size()&&nums[level]==nums[level+1])level++;
        DFS(res, out, nums, level+1);
	}
}
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
	    vector<int> out;
	    DFS(res, out, nums,0);
	    return res;
    }