401. Binary Watch

A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent theminutes (0-59).

Each LED represents a zero or one, with the least significant bit on the right.

For example, the above binary watch reads "3:25".

Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.

Example:

Input: n = 1Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]

Note:

• The order of output does not matter.
• The hour must not contain a leading zero, for example "01:00" is not valid, it should be "1:00".
• The minute must be consist of two digits and may contain a leading zero, for example "10:2" is not valid, it should be "10:02".

嗯， 一道easy我还看了答案……丢人。

两种方法，第一种是用dfs回溯做，第二种是用bitset做，bitset明显快很多很多很多……

class Solution {public:    vector<int> hour = {1, 2, 4, 8}, minutes = {1, 2, 4, 8, 16, 32};    vector<string> readBinaryWatch(int num) {                vector<string> res;        dfs(res, make_pair(0, 0), num, 0);        return res;    }        void dfs(vector<string>& res, pair<int, int> t, int num, int index){        if(num == 0){            res.push_back(to_string(t.first) + (t.second >= 10 ? ":" : ":0") + to_string(t.second));            return ;        }                for(int i = index; i != 10; ++ i){            if(i < 4){                t.first += hour[i];                if(t.first >= 12) i = 3;//剪枝                else dfs(res, t, num - 1, i + 1);                t.first -= hour[i];            }            else{                t.second += minutes[i - 4];                if(t.second >= 60) break;//剪枝                else dfs(res, t, num - 1, i + 1);                t.second -= minutes[i - 4];            }        }    }};

class Solution {public:    vector<int> hour = {1, 2, 4, 8}, minutes = {1, 2, 4, 8, 16, 32};    vector<string> readBinaryWatch(int num) {                vector<string> res;        for(int h = 0; h < 12; ++ h)            for(int m = 0; m < 60; ++ m){                if(bitset<4>(h).count() + bitset<6>(m).count() == num)                    res.push_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m));            }                        return res;    }    };