[leetcode]401. Binary Watch

最新推荐文章于 2025-07-16 16:22:45 发布

Brett-Xu

最新推荐文章于 2025-07-16 16:22:45 发布

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分类专栏： LeetCode 文章标签： LeetCode C++

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本文介绍了一种特殊的二进制手表，通过顶部的4个LED灯表示小时（0-11），底部的6个LED灯表示分钟（0-59）。文章详细解释了如何根据LED灯亮起的数量，计算出所有可能显示的时间，并提供了一个C++实现的解决方案。

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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".

分析：

一个二进制手表，第一行四个灯表示时，下面六个灯表示分，要求根据灯亮的个数返回所有可能的时间，注意时前面不添0，分保证是2位组成，不满2位用0补。时针从0遍历到11，分针从0遍历到59，然后把时针的数组左移6位加上分针的数值，然后统计1的个数，即为亮灯的个数，遍历所有的情况，当其等于num的时候，存入结果res中。

class Solution {
public:
    vector<string> readBinaryWatch(int num) {
        int h = 0;
        int m = 0;
        vector<string> res;
        for(h=0; h<12; h++)
        {
            for(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;
    }
};