leetcode 1185. Day of the Week 解法 python

一.问题描述

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"}.

 

Example 1:

Input: day = 31, month = 8, year = 2019
Output: "Saturday"

Example 2:

Input: day = 18, month = 7, year = 1999
Output: "Sunday"

Example 3:

Input: day = 15, month = 8, year = 1993
Output: "Sunday"

 

Constraints:

• The given dates are valid dates between the years 1971 and 2100.

二.解决思路

这道题我真的不知道咋说，要水也不能水成这个样子把。。。

主要就注意以下从啥时候开始计算

起始点是1971年1月1日，这天是周五

之后按一年年计算离给定日子总共有多少天，最后取余就好

涉及到闰年的判断：

闰年：年数能整除4， 当年数能整除100的时候还必须整除400，不然不算闰年。

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三.源码

class Solution:
    def dayOfTheWeek(self, day: int, month: int, year: int) -> str:
        # 1971 1.1 is Friday
        num2day={0:'Thursday',1:'Friday',2:'Saturday',3:'Sunday',4:'Monday',5:'Tuesday',6:'Wednesday'}
        mon2num=[0,31,28,31,30,31,30,31,31,30,31,30,31]
        count=0
        for i in range(1971,year):  
            if i%400==0 or (i%4==0 and i%100!=0):
                count+=366
            else :
                count+=365
        if year%400==0 or (year%4==0 and year%100!=0):
                mon2num[2]+=1
        return num2day[(count+sum(mon2num[:month])+day)%7]