1185. Day of the Week

Given a date, return the corresponding day of the week for that date.

The input is given as three integers representing the day, month and year respectively.

Return the answer as one of the following values {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}.

这题有三种解法，

一是 利用Zeller Formula去计算，这个公式是什么感兴趣自己查。我没兴趣

二是知道1971-01-01这天是星期几，然后计算给出的时间到整天的天数，%7就得到是星期几

三是不需要知道1971-01-01是星期几，只需要知道今天是星期几，然后计算今天到1971-01-01的天数，目标日期到1971-01-01的天数，两个天数做个diff 然后% 7，就能知道是星期几了。

class Solution(object):
    def dayOfTheWeek(self, day, month, year):
        """
        :type day: int
        :type month: int
        :type year: int
        :rtype: str
        """
        def hasLeapDay(year):
            return 1 if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0 else 0

        # 2020 07 01
        dayNames = ["Wednesday", "Thursday", "Friday", "Saturday", "Sunday", "Monday", "Tuesday"]
        
        daysInMonth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
        # days since 31, 12, 1970
        def daysSinceStart(day, month, year):
            numDays = 0
            for y in range(year - 1, 1970, -1):
                numDays += 365 + hasLeapDay(y)
            numDays += sum(daysInMonth[:month-1])
            numDays += day 
            if month > 2:    
                numDays += hasLeapDay(year)
            return numDays

        knownStart = daysSinceStart(1,7,2020)
        d = daysSinceStart(day, month, year) 
        return dayNames[ (d - knownStart) % 7]