90. Subsets II

1.問題

2.想法

• 提問

  • 確認題意: 求數列所有的可能組合, 數列是未經排序, 而且可能有重複數字

• function header, parameter

• test input

• 說明想法

  • 做法跟原本的78.subset相同, 但是需要先排序, 而且若num[i] == num[i - 1]時則略過

• 測試計算複雜度

3.程式碼

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> record;
        vector<bool> used(nums.size(), false);
        sort(nums.begin(), nums.end());
        for (int i = 1; i <= nums.size(); i++) {
            getSubsets(nums, res, i, 0, record, used);
        }
        res.push_back(vector<int>());
        return res;
    }
private:
    void getSubsets(vector<int>& nums, vector<vector<int>>& res, int n, int start, vector<int>& record, vector<bool>& used) {
        if (record.size() == n) {
            res.push_back(record);
            return;
        }
        
        for (int i = start; i < nums.size(); i++) {
            if (!used[i]) {
                if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {
                    continue;
                }
                record.push_back(nums[i]);
                used[i] = true;
                getSubsets(nums, res, n, i + 1, record, used);
                record.pop_back();
                used[i] = false;
            }
        }
    }
};