401 binary watch
401. Binary Watch
题目: https://leetcode.com/problems/binary-watch/
难度: Easy
思路：
一看到位操作，我的内心是拒绝的。
我也有想这样的想法，因为其实可以的组合并没有那么多，干脆枚举算了，然而也没有动手来写，直到被发了题解的截屏。
class Solution(object):
def readBinaryWatch(self, num):
"""
:type num: int
:rtype: List[str]
"""
hour = { 0 : ['0'],
1:['1','2','4','8'],
2:['3','5','6','9','10'],
3:['7','11']
}
minute = { 0:['00'],
1: ['01','02','04','08','16','32'],
2: ['03','05','06','09','10','12','17','18','20','24','33','34','36','40','48'],
3: ['07','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']
}
res = []
#num = num for hour + num for minute
i = 0
while i <= 3 and i <= num:
if num - i <= 5:
for str1 in hour[i]:
for str2 in minute[num-i]:
res.append(str1 + ':' + str2)
i += 1
return res
关于循环那处，因为hour的led最多只有4个，所以这样写循环
