【DFS】LeetCode 40. Combination Sum II

Solution1：我的答案
同39题。DFS时间复杂度$O(2^n)$，空间复杂度$O(kn)$，k是最终答案的数量，n是元素个数
去重，用set，比较偷懒的做法

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        set<vector<int> > res;
        vector<int> temp;
        sort(candidates.begin(), candidates.end());
        int start = 0;
        my_combinate(res, temp, candidates, target, start);
        return vector<vector<int> > (res.begin(), res.end());
    }

    void my_combinate(set<vector<int> >& res, vector<int>& temp,
                      vector<int>& candidates, int target, int start) {
        if (target < 0) return;
        else if (target == 0)
            res.insert(temp);
        else {
            for (int i = start; i < candidates.size(); i++) {
                temp.push_back(candidates[i]);
                my_combinate(res, temp, candidates, target - candidates[i], i + 1);
                temp.pop_back();
            }
        }
    }
};

Solution2：
去重：同一层不利用相同的数

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int> > res;
        vector<int> temp;
        sort(candidates.begin(), candidates.end());
        int start = 0;
        my_combinate(res, temp, candidates, target, start);
        return res;
    }

    void my_combinate(vector<vector<int> >& res, vector<int>& temp,
                      vector<int>& candidates, int target, int start) {
        if (target < 0) return;
        else if (target == 0)
            res.push_back(temp);
        else {
            for (int i = start; i < candidates.size(); i++) {
                if (i > start && candidates[i] == candidates[i-1]) continue;//同一层不用相同的数
                temp.push_back(candidates[i]);
                my_combinate(res, temp, candidates, target - candidates[i], i + 1);
                temp.pop_back();
            }
        }
    }
};