401. Binary Watch

最新推荐文章于 2024-10-13 00:45:51 发布

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最新推荐文章于 2024-10-13 00:45:51 发布

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分类专栏： leetcode 文章标签：位运算

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本文介绍了一种特殊的手表——二进制手表的工作原理及如何根据点亮的LED数量读取时间。通过遍历所有可能的时间组合，找出符合特定LED数量亮起的情况。

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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".

小时数和分钟数用二进制表示，要求给出亮n个灯（即二进制数的1的位数为n）能代表哪些时间。遍历全部时间（0-12小时，每个小时的0-60分钟），如果一个时间的二进制数1的位数为n即满足条件，把它加在结果后面。统计一个时间二进制数1的位数的方法是利用"与"操作，不断清除二进制数表示中最右边的1，同时累加计数器，直至二进制数为0。

代码：

class Solution {
public:
	vector<string> readBinaryWatch(int num)
	{
		vector<string> res;
		for(int i = 0; i < 12; ++i)
		{
			for(int j = 0; j < 60; ++j)
			{
				if(count1(i << 6 | j) == num)
				{
					res.push_back(to_string(i) + (j < 10 ? ":0" : ":") + to_string(j));
				}
			}
		}
		return res;
	}
private:
	int count1(int n)
	{
		int cnt = 0;
		while(n)
		{
			n &= (n - 1);
			++cnt;
		}
		return cnt;
	}
};