LeetCode401 Binary Watch

文章目录
  1. 1. 描述
  2. 2. 样例
  3. 3. 思路
  4. 4. 代码

描述

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.

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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.

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”.

样例

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Input: n = 1
Return: ["1:00", "2:00", "4:00", "8:00", "0:01",
"0:02", "0:04", "0:08", "0:16", "0:32"]

思路

这是一个神奇的二进制手表… 问当有n个LED灯亮起时,可能的时间是多少。

直接枚举所有时间,再判断当前时间的二进制表示中1的个数是否满足即可。用到了 一个小技巧就是把时间用二进制压缩表示。h << 6 | m,h是小时数,m是分钟数,由于分钟数最多用6个二进制位表示即可,所以低6位用来表示分钟,高位则表示小时。最后再用系统自带的__builtin_popcount()函数计算二进制中1的个数。

代码

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