Binary Watch

题目要求如下：

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".

解法一：

解题思路：将每一个小时中1出现的次数记录下来，比如3点，二进制是11,1出现了两次，用该位数作为key，把3转化成string记录进去；分钟类似；然后遍历num进行查找即可。

代码如下：

class Solution {public:    int nums_1(int n){        int num = 0;        while(n){            num++;            n &= n-1;        }        return num;    }    vector<string> readBinaryWatch(int num) {        //整数存储亮的灯的个数，vector存储亮这些灯可能代表的时间的字符串        unordered_map<int,vector<string>> hours;        unordered_map<int,vector<string>> minus;        vector<string> result;                for(int i = 0 ; i<12 ; i++){            hours[nums_1(i)].push_back(to_string(i));        }        for(int i = 0 ; i<10 ; i++){            minus[nums_1(i)].push_back("0"+to_string(i));        }        for(int i = 10; i<60 ; i++){            minus[nums_1(i)].push_back(to_string(i));        }                for(int i = 0; i<=3 && i<= num ; i++){            if(num - i >=6) continue;            int len_h = hours[i].size();            int len_m = minus[num-i].size();            for(int j = 0; j < len_h ; j++){                for(int k = 0; k < len_m; k++){                    string temp = hours[i][j] +":"+minus[num-i][k];                    result.push_back(temp);                }             }        }        return result;    }};

解法二：

解题思路：使用bitset统计hour和minute中出现1的此时的和，判断是否和num相等，如果相等，则放入结果中。

vector<string> readBinaryWatch(int num) {    vector<string> rs;    for (int h = 0; h < 12; h++)    for (int m = 0; m < 60; m++)        if (bitset<10>(h << 6 | m).count() == num)            rs.emplace_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m));    return rs;}

解法三：回溯法