401. Binary Watch

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.

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".

思路：

题目考察二进制表，4个时间灯，6个分钟灯，给出亮灯的个数求所有可能组成的时间。我们把亮着的比喻成二进制1，灭着的为0.

这样我们计算时间的二进制为1的个数加上分钟时间二进制为1的个数，当个数和等于num时则为符合规则的情况。

代码：

	public List<String> readBinaryWatch(int num) {
		List<String> res = new ArrayList<>();
		for (int h = 0; h < 12; h++) {
			for (int m = 0; m < 60; m++) {
				int total = bitCount(h) + bitCount(m);
//				也可以直接使用Integer自带接口计算1的个数
//				int total = Integer.bitCount(m) + Integer.bitCount(h);
				if (total == num) {
					String s = "";
					s += h + ":";
					s += (m < 10) ? "0" + m : m;
					res.add(s);
				}
			}
		}
		return res;
	}

	public int bitCount(int num) {
		int res = 0;
		while (num > 0) {
			if ((num & 1) == 1) {
				res++;
			}
			num >>= 1;
		}
		return res;
	}