Share my recursive and bit operation solutions


  • 2
    S
    // bit operation: first get all results, finally delete the duplicates
    vector<vector<int> > subsetsWithDup1(vector<int> &S) {
        sort(S.begin(), S.end());
        int len = S.size();
        int n = pow(2, len);
        vector<vector<int> > ans(n, vector<int>());
        
        for (int i = 0; i < len; ++i) {
            for (int j = 0; j < n; ++j) {
                if (j >> i & 1) ans[j].push_back(S[i]);
            }
        }
        
        sort(ans.begin(), ans.end());
        ans.resize(unique(ans.begin(), ans.end()) - ans.begin());
        return ans;
    }
    
    // iterative solution: ignore the duplicates directly
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        sort(S.begin(), S.end());
        vector<vector<int> > ans(1, vector<int>());
        int len = S.size();
        int s = 0, e = 0;
        
        vector<int> tmp;
        for(int i = 0; i < len; ++i) {
            s = i >= 1 && S[i] == S[i-1] ? e : 0;
            e = ans.size();
            for(int j = s; j < e; ++j) {
                tmp = ans[j];
                tmp.push_back(S[i]);
                ans.push_back(tmp);
            }
        }
        
        return ans;
    }

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