39. Combination Sum

注意，sort的时候，千万不要直接改变list,要用临时数组来代替；因为递归过程中后面还要pop_back;

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        set<vector<int>> myres;
        if(candidates.empty()) return result;
        vector<int> list;
        dfs(candidates, myres, list, target);
        for(auto iter = myres.begin(); iter != myres.end(); iter++){
            result.push_back(*iter);
        }
        return result;
        
    }
private:
    void dfs(vector<int>& candidates, set<vector<int>>& myres, vector<int>& list, int now){
        if(now < 0) return;
        if(now == 0){
            vector<int> tmp(list);
            sort(tmp.begin(), tmp.end());//不能改变list, 要用另外的临时数组
            myres.insert(tmp);
            return;
        }
        for(int i = 0; i < candidates.size(); i++){
            if(now < candidates[i] ) continue;
            list.push_back(candidates[i]);
            dfs(candidates, myres, list, now - candidates[i]);
            list.pop_back();
        }
    }
};

解法二：
用一个标记，这样可以防止重复的答案，比如[2,2,3] 和[2,3,2]

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> result;
        vector<int> list;
        if(candidates.empty()) return result;
        sort(candidates.begin(), candidates.end());
        dfs(candidates, result, list, target, 0);
        return result;
        
    }
private:
    void dfs(vector<int> candidates, vector<vector<int>>& result, vector<int>& list, int cur, int begin){
        if(cur == 0){
            result.push_back(list);
            return;
        }
        for(int i = begin ; i < candidates.size() && candidates[i] <= cur; i++){
            list.push_back(candidates[i]);
            dfs(candidates, result, list, cur - candidates[i],i);//因为这个值可以重复用，一定要注意，既然在了循环里，所以要用的是i,这样可以防止重复的答案。
            list.pop_back();
        }
    }
};