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).

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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: 

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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”.

思路

一开始是懵的……没做出来。

先尝试最直白的思路，做位运算。

当num=1的时候，做一个循环，把左移1位至10位的结果都放到结果的list里面，之后再对结果的list进行格式转换就好。一共10位，高四位是小时数，低六位是分钟数。num=2的时候，做两个循环以此类推。但是不知道如何利用nums来控制循环的层数，于是卡在这个思路里面没有做出来。

但是后来才发现，上面的思路就是排列组合当中的组合数。所以看到了有些答案用了itertools里面combinations的方法。也是极好的。实际上，之前卡住的时候还不会用递归求解组合数，也可以在这里利用递归的方法求组合数。

最好的一种方法是反推法。把所有的12×60种时间全部遍历一遍。在遍历的过程中，如果bits的个数为nums，就把它记录在结果列表里面。 穷举法就不多说了，很暴力，但是也很有效。

贴一个后来写的代码，利用的是穷举法，暴力求解。

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class Solution(object):
    def readBinaryWatch(self, num):
        """
        :type num: int
        :rtype: List[str]
        """
        res = []
        for i in range(12):
            for j in range(60):
                if (self.numofbin(i) + self.numofbin(j) == num):
                    hour = str(i)
                    minute = str(j)
                    if (j < 10):
                        minute = "0" + minute
                    res.append(hour+":"+minute)
        return res
    def numofbin(self, num):
        count = 0
        while (num):
            count += num % 2
            num = num / 2
        return count