LeetCode 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 the minutes (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 = 1
Return: ["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".

DFS，先枚举灯的个数，然后根据个数搜索。

class Solution {public:    int re[6] = {1, 2, 4, 8, 16, 32};    int hours[4] = {0, 0, 0, 0};    int minutes[6] = {0, 0, 0, 0, 0, 0};    vector<string> v;    void dfsh(int start, int numh, int numm){        if(start == 4 && numh) return;         if(!numh){            dfsm(0, numm);        }else{            hours[start] = 1;            dfsh(start + 1, numh - 1, numm);            hours[start] = 0;            dfsh(start + 1, numh, numm);        }    }    void dfsm(int start, int numm){        if(start == 6 && numm) return;         if(!numm){            int h = 0, m = 0, i;            for(i = 0; i < 4; i ++){                if(hours[i]) h += re[i];            }            for(i = 0; i < 6; i ++){                if(minutes[i]) m += re[i];            }            string tmp = "";            if(h > 11 || m > 59) return;            tmp = to_string(h) + ":";            if(m < 10) tmp += "0" + to_string(m);            else tmp += to_string(m);            v.push_back(tmp);        }else{            minutes[start] = 1;            dfsm(start + 1, numm - 1);            minutes[start] = 0;            dfsm(start + 1, numm);        }    }    vector<string> readBinaryWatch(int num) {        int i;        for(i = 0; i <= 4; i ++){            if(num - i <= 6 && num - i >= 0){                dfsh(0, i, num - i);            }        }        return v;    }};