Mock 002 Print 5-min intervals

Input - (“mon 10:00 am”, mon 11:00 am)
Output - [11005, 11010, 11015…11100]
Output starts with 1 if the day is monday, 2 if tuesday and so on till 7 for sunday
Append 5 min interval times to that till the end time
So here it is 10:05 as first case, so its written as 11005
2nd is 10:10 so its written as 11010

题目大意：

DD的面经题，给定开始时间和结束时间，求5分钟的间隔时间，注意要round to 5min

解题思路：

由于非10进制，所以开一个类来计算进制

解题步骤：

N/A

注意事项：

1. 实现lt函数
2. 12am, 12pm要mod 12
3. (0 if parts[2] == ‘am’ else 12)加括号
4. 开始时间到到5分钟端点，结束时间加1分钟，由于只实现了lt

Python代码：

DAY_DICT = {'mon': 1, 'tue': 2, 'wed': 3, 'thu': 4, 'fri': 5, 'sat': 6, 'sun': 7}
class Solution(TestCases):

    def get_intervals(self, start, end) -> List:
        start_time = Time(start)
        end_time = Time(end)
        if start_time.min % 5 > 0:
            start_time.add(5 - start_time.min % 5)
        end_time.add(1)
        res = []
        while start_time < end_time:
            res.append(start_time.get_numeric())
            start_time.add(5)
        return res


class Time:

    def __init__(self, time):
        parts = time.split(' ')
        day = DAY_DICT[parts[0]]
        time_parts = parts[1].split(':')
        hour = int(time_parts[0]) % 12 + (0 if parts[2] == 'am' else 12) # remember paren (0 ...12), and % 12
        self.day = day
        self.hour = hour
        self.min = int(time_parts[1])

    def __lt__(self, other):
        if self.day < other.day or (self.day == other.day and self.hour < other.hour) or \
                (self.day == other.day and self.hour == other.hour and self.min < other.min):
            return True
        else:
            return False

    def get_numeric(self):
        return self.day * 10000 + self.hour * 100 + self.min

    def add(self, mins):
        self.min += mins
        if self.min == 60:
            self.min = 0
            self.hour += 1
        if self.hour == 24:
            self.hour = 0
            self.day += 1
        if self.day == 7:
            self.day = 0

算法分析：

时间复杂度为O(n)，空间复杂度O(1)