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".
Subscribe to see which companies asked this question.
思路:用的暴力求解,就是从最大往后减,如果1的个数等于所给数据,则添加。
public class Solution {
public List<String> readBinaryWatch(int num) {
int hour = 11;
int minute = 59;
String t = "";
int one = 0;
List<String> list = new ArrayList<>();
if (num > 8 || num < 1) {
if (num == 0) {
t = "0:00";
list.add(t);
return list;
}
return list;
}
while (hour >= 0) {
one = Integer.bitCount(hour) + Integer.bitCount(minute);
t = "";
if (one == num) {
t = t + hour;
if (minute < 10) {
t = t + ":0" + minute;
} else {
t = t + ":" + minute;
}
list.add(new String(t));
}
one = 0;
minute--;
if (minute < 0) {
minute = 59;
hour--;
}
}
return list;
}
}