[leetcode]401. Binary Watch
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;
}
};