leetcode1185(一周中的第几天)

一周中的第几天

给你一个日期，请你设计一个算法来判断它是对应一周中的哪一天。

输入为三个整数：day、month 和 year，分别表示日、月、年。

您返回的结果必须是这几个值中的一个 {“Sunday”, “Monday”, “Tuesday”, “Wednesday”, “Thursday”, “Friday”, “Saturday”}。

示例 1：

输入：day = 31, month = 8, year = 2019
输出：“Saturday”
示例 2：

输入：day = 18, month = 7, year = 1999
输出：“Sunday”
示例 3：

输入：day = 15, month = 8, year = 1993
输出：“Sunday”

提示：

给出的日期一定是在 1971 到 2100 年之间的有效日期。

*/C++解法
class Solution {
public:
    string dayOfTheWeek(int day, int month, int year) {
        int days = 5;
        string week[7] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
        for (int i = 1971; i < year; i++) {
            if((i%400==0) || (i%4==0 && i%100!=0)) {
                days+=366;
            }else {
            days+=365;
            }
        }
        int m[12] ={31,28,31,30,31,30,31,31,30,31,30,31};
        for (int i = 0; i <month -1; i++){
            days+=m[i];
        }
        days += day -1;
        if(((year%400==0) || (year%4==0 && year%100!=0)) && month>=3)
            return week[(days+1)%7];
        else
            return week[days%7];
    }
};


*/Python解法
class Solution:
    def dayOfTheWeek(self, day: int, month: int, year: int) -> str:
        from datetime import datetime
        days = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]
        return days[datetime(year, month, day).weekday()]
        
*/
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]