[LeetCode] Binary Watch

Posted 2020-09-28 immjc

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".

给定一个二进制手表，如上图有10个LED组成。给定一个数字代表点亮的LED个数。把亮着的LED所代表的时间返回一个字符串数组。Note中要求了时间输出的格式。

h << 6 表示h的二进制表示左移6位。

h << 6 | m表示左移后的二进制与m取或，即h末尾的二进制由m的二进制位代替。

bitset<10>(h << 6 | m)表示一个10位二进制数，h的二进制前面几位取0构成一个10位二进制数。10个LED。

count()函数返回bitset<10>(h << 6 | m)中为1的个数。

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.emplace_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m));
        return res;
    }
};
// 3 ms