[leetcode: Python]401. Binary Watch

Title:

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”.

方法一：性能45ms

枚举小时h和分钟m，然后判断二进制1的个数是否等于num

class Solution(object):
def readBinaryWatch(self, num):
"""
:type num: int
:rtype: List[str]
"""
res = []
for h in range(12):
for m in range(60):
if (bin(h)+bin(m)).count('1') == num:
res.append('%d:%02d' % (h, m))
return res

方法二：35ms

class Solution(object):
def readBinaryWatch(self, num):
"""
:type num: int
:rtype: List[str]
"""
# hh_summary, mm_summary = self.get_summary()
hh_summary = {
0: [0],
1: [1, 2, 4, 8],
2: [3, 5, 6, 9, 10],
3: [7, 11],
}
mm_summary = {
0: [0],
1: [1, 2, 4, 8, 16, 32],
2: [3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48],
3: [7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 35, 37, 38, 41, 42, 44,
49, 50, 52, 56],
4: [15, 23, 27, 29, 30, 39, 43, 45, 46, 51, 53, 54, 57, 58],
5: [31, 47, 55, 59],
}

valid_time = []
for hh in range(num + 1):
mm = num - hh
valid_hh = hh_summary.get(hh, [])
valid_mm = mm_summary.get(mm, [])
for h in valid_hh:
for m in valid_mm:
valid_time.append('{0}:{1:02d}'.format(h, m))
return valid_time

# def get_summary(self):
#     hh_summary = self._get_summary(11)
#     mm_summary = self._get_summary(59)
#     return hh_summary, mm_summary

# def _get_summary(self, num):
#     summary = {}
#     for n in range(num + 1):
#         bit = self.get_bit(n)
#         summary[bit] = summary.get(bit, []) +

#     return summary

# def get_bit(self, n):
#     bit = 0
#     while n > 0:
#         bit += n & 1
#         n >>= 1
#     return bit