Binary Watch二进制时间

［抄题］：

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".

[暴力解法]：

时间分析：n2

空间分析：

［思维问题］：

不知道和回溯法有什么关系。一看特别麻烦，果断用暴力解法了

［一句话思路］：

用bitCount转化为二进制数

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

1. 表的小时不超过12
2. 用String.format严格控制字符串格式

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

［复杂度］：Time complexity: O(n2) Space complexity: O(n)

［英文数据结构或算法，为什么不用别的数据结构或算法］：

麻烦

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

[代码风格] ：

public class Solution {
    /*
     * @param : the number of "1"s on a given timetable
     * @return: all possible time
     */
    public List<String> readBinaryWatch(int num) {
        List<String> time = new ArrayList<String>();
        for (int h = 0; h < 12; h++) {//12 not 24
            for (int m = 0; m < 60; m++) {
                if (Integer.bitCount(h) + Integer.bitCount(m) == num) {
                    time.add(String.format("%d:%02d", h, m));//String's strict format
                }
            }
        }
        return time;
    }
};