2.1.2.一串數列中取n個數字, 共有幾種組合

• example code

            class Solution {
                vector<int> hour = {1, 2, 4, 8};
                vector<int> minute = {1, 2, 4, 8, 16, 32};
            public:
                vector<string> readBinaryWatch(int num) {
  
                    vector<string> res;
                    //初始值: time = 00:00, start point = 0
                    helper(res, make_pair(0, 0), num, 0);
                    return res;
                }
                //res: 儲存字串
                //time: 累積時間
                //num: 剩餘數量
                //start_point: 數量起始值
                void helper(vector<string>& res, pair<int, int> time, int num, int start_point){
                    //終止條件: 剩餘數量=0
                    if(num == 0){
                        res.push_back(to_string(time.first) + (time.second < 10 ? ":0" : ":") + to_string(time.second));
                        return;
                    }
                    for(int i = start_point; i < hour.size() + minute.size(); i++){
                        if(i < hour.size()){
                            //hour
                            time.first = time.first + hour[i];
                            if(time.first < 12){
                                helper(res, time, num - 1, i + 1);
                            }
                            time.first = time.first - hour[i];
                        }else{
                            //minute
                            time.second = time.second + minute[i - hour.size() ];
                            if(time.second < 60){
                                helper(res, time, num - 1, i + 1);
                            }
                            time.second = time.second - minute[i - hour.size() ];
                        }
                    }
                }
            };