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.

Binary Watch

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:

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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”.

方法1：回溯法
这是一个典型的回溯法的题目，其实就是在10个位置中选n个，前6位组成Minute，后4位组成Hour.
代码如下：

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public List<String> readBinaryWatch(int num) {
    List<String> ans = new ArrayList<>();
    backtracking(num, 0, 0, 0, ans);
    return ans;
}
private void backtracking(int num, int start, int res, int count, List<String> ans) {
    if (count == num) {
        String str = generateTime(res);
        if (str != null)
            ans.add(str);
        return;
    }
    for (int i = start; i < 10; i++) {
        res |= 1 << i;
        count++;
        backtracking(num, i + 1, res, count, ans);
        int mask = 1 << i;
        mask = ~mask;
        res &= mask;
        count--;
    }
}
private String generateTime(int res) {
    int minute = res & 0x3f;
    int hour = res >> 6;
    if (hour >= 12 || minute >= 60)
        return null;
    return hour + ":" + (minute >= 10 ? minute : "0" + minute);
}

方法2：排除法
Ref: https://discuss.leetcode.com/topic/59374/simple-python-java
找到所有Hour和Minute的组合，检查bit count是不是等于num，如果相等，就是所要的结果

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public List<String> readBinaryWatch(int num) {
    List<String> times = new ArrayList<>();
    for (int h = 0; h < 12; h++)
        for (int m = 0; m < 60; m++)
            if (Integer.bitCount(h * 64 + m) == num)
                times.add(String.format("%d:%02d", h, m));
    return times;
}

这种做法写起来简单，但是时间复杂度不如第一种好，而且这个题目的考点应该就是回溯法，用第二种方法应该不能过关。

另外:
h + ":" + (m >= 10 ? m : "0" + m) 比 String.format("%d:%02d", h, m) 快上不少。