401. Binary Watch | 二进制手表

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".

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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;}}