leetcode 1185. Day of the Week 解法 python

日期转星期算法解析

最新推荐文章于 2024-03-23 17:40:23 发布

原创最新推荐文章于 2024-03-23 17:40:23 发布 · 973 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #python #算法 #编程 #闰年

leetcode算法从零到结束专栏收录该内容

112 篇文章

订阅专栏

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。

更多leetcode算法题解法请关注我的专栏leetcode算法从零到结束或关注我

欢迎大家一起套路一起刷题一起ac

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]