LeetCode-Backtracking-401. Binary Watch

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问题：https://leetcode.com/problems/binary-watch/?tab=Description
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”.
用表上的10个数字表示时间，上面的四个代表小时，下面的六个代表分钟。
分析：对小时h和分钟m进行遍历，总共有12*60种可能（分别为12小时和60分钟），之后将小时h右移6位（右边6位为分钟m的二进制，另外的4位代表小时h的二进制）转换成题目要求的10个二进制，并调用bitset的count计算二进制中1的个数，若1的个数等于num，则该时间满足题目要求。再把显示的时间改成所要求的时间格式。
参考C++代码：

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

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