目录
Day of the Week
题目大意
常规方法
Python代码
Golang代码
C++代码
基姆拉尔森公式
Python代码
Golang代码
C++代码
使用库函数
Python代码
Golang代码
C++代码
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”} .
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”
题目大意
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”}。
提示：
给出的日期一定是在 1971 到 2100 年之间的有效日期。
解题思路：
给出一个日期，要求算出这一天是星期几。
常规方法
从1971.1.1起，先累计整年year、整月month-1的天数，再加上最后一个月month的天数day，然后总天数减1后与7求余。最后得到的余数在星期字串数组中位置索引，显然前提要知道1971.1.1这个基准日期是星期几，再作一个索引位移就是答案。
另外常规方法还需要判断year是否闰年，规则：y%4==0 and y%100!=0 or y%400==0，据说是1582
Python代码
python代码非常简单，不需另外导入库只用内置函数就能搞定。
class Solution(object):def DayOfWeek(self, year, month, day):days = 0isLeapYear = lambda y:y%4==0 and y%100!=0 or y%400==0monthday = [31,28,31,30,31,30,31,31,30,31,30,31]week = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"]monthday[1] = 29 if isLeapYear(year) else 28for i in range(1971,year):days += 366 if isLeapYear(i) else 365days += sum(monthday[:month-1], day-1)return week[(days+5)%7]if __name__ == "__main__":s = Solution()print(s.DayOfWeek(2019,8,31))print(s.DayOfWeek(1999,7,18))print(s.DayOfWeek(1993,8,15))print(s.DayOfWeek(1971,6,12))print(s.DayOfWeek(2023,2,22))print(s.DayOfWeek(2040,6,13))输出：
Saturday
Sunday
Sunday
Saturday
Wednesday
Wednesday
Golang代码
基本原理相同，另外自定义一个数组求和公式即可。
package mainimport "fmt"func DayOfWeek(year int, month int, day int) string {days := 0isLeapYear := func(y int) bool {return y%4 == 0 && y%100 != 0 || y%400 == 0}Sum := func(nums []int, initNum int) int {var sumNum int = 0for _, num := range nums {sumNum += num}return sumNum + initNum}monthday := []int{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}week := []string{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}if isLeapYear(year) {monthday[1] = 29} else {monthday[1] = 28}for i := 1971; i < year; i++ {if isLeapYear(i) {days += 366} else {days += 365}}days += Sum(monthday[:month-1], day-1)return week[(days+5)%7]}func main() {fmt.Println(DayOfWeek(2019, 8, 31))fmt.Println(DayOfWeek(1999, 7, 18))fmt.Println(DayOfWeek(1993, 8, 15))fmt.Println(DayOfWeek(1971, 6, 12))fmt.Println(DayOfWeek(2023, 2, 22))fmt.Println(DayOfWeek(2040, 6, 13))}输出：
Saturday
Sunday
Sunday
Saturday
Wednesday
Wednesday
成功: 进程退出代码 0.
C++代码
引入C++11的容器vector，可以省掉最后一个非整年的各月份日数循环累加，只要用库中的函数accumulate，方便累加非整年的各月份日数，并且把day作为基准数一并累加掉。
#include#include#includeusing namespace std;class Solution{public:string DayOfWeek(int year, int month, int day){int days = 0;auto isLeapYear = [](int y) { return y%4==0 && y%100!=0 || y%400==0; };vector monthday = {31,28,31,30,31,30,31,31,30,31,30,31};vector week = {"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};monthday[1] = isLeapYear(year) " />基姆拉尔森公式 万能的日期计算公式，不用知道基准日是哪一天，也不需要判断year是否为闰年。
公式：weekday = (day+2month+3(month+1)/5+year+year/4-year/100+year/400+1)%7
注意：1月和2月需看做上一年的13月与14月，即 month<3时，year-=1; month+=12
Python代码
class Solution(object):def DayOfWeek(self, year, month, day):week = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"]if month<3: year, month = year-1, month+12weekday = (day+2*month+3*(month+1)//5+year+year//4-year//100+year//400+1)%7return dict(zip(range(7),week)).get(weekday)if __name__ == "__main__":s = Solution()print(s.DayOfWeek(2019,8,31))print(s.DayOfWeek(1999,7,18))print(s.DayOfWeek(1993,8,15))print(s.DayOfWeek(1971,6,12))print(s.DayOfWeek(2023,2,22))print(s.DayOfWeek(2040,6,13))
Golang代码
package mainimport "fmt"func DayOfWeek(year int, month int, day int) string {week := []string{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}if month < 3 {year -= 1month += 12}weekday := (day + 2*month + 3*(month+1)/5 + year + year/4 - year/100 + year/400 + 1) % 7return week[weekday]}func main() {fmt.Println(DayOfWeek(2019, 8, 31))fmt.Println(DayOfWeek(1999, 7, 18))fmt.Println(DayOfWeek(1993, 8, 15))fmt.Println(DayOfWeek(1971, 6, 12))fmt.Println(DayOfWeek(2023, 2, 22))fmt.Println(DayOfWeek(2040, 6, 13))}
C++代码
#includeusing namespace std;class Solution{public:string DayOfWeek(int year, int month, int day){const char *week[7] = {"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};if (month < 3) {year -= 1;month += 12;} int weekday = (day+2*month+3*(month+1)/5+year+year/4-year/100+year/400+1)%7;return week[weekday];}};int main(){Solution s;cout << s.DayOfWeek(2019,8,31) << endl;cout << s.DayOfWeek(1999,7,18) << endl;cout << s.DayOfWeek(1993,8,15) << endl;cout << s.DayOfWeek(1971,6,12) << endl;cout << s.DayOfWeek(2023,2,22) << endl;cout << s.DayOfWeek(2040,6,13) << endl;return 0;}
使用库函数
Python代码
datetime库
import datetimeclass Solution(object):def DayOfWeek(self, year, month, day):week = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"]weekday = datetime.date(year,month,day).isoweekday()return week[weekday%7]if __name__ == "__main__":s = Solution()print(s.DayOfWeek(2019,8,31))print(s.DayOfWeek(1999,7,18))print(s.DayOfWeek(1993,8,15))print(s.DayOfWeek(1971,6,12))print(s.DayOfWeek(2023,2,22))print(s.DayOfWeek(2040,6,13))
calendar库
import calendarclass Solution(object):def DayOfWeek(self, year, month, day):week = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"]weekday = calendar.weekday(year,month,day)+1return week[weekday%7]if __name__ == "__main__":s = Solution()print(s.DayOfWeek(2019,8,31))print(s.DayOfWeek(1999,7,18))print(s.DayOfWeek(1993,8,15))print(s.DayOfWeek(1971,6,12))print(s.DayOfWeek(2023,2,22))print(s.DayOfWeek(2040,6,13))
Golang代码
time库，超级省事，连星期数组都不用了。
package mainimport ("fmt""time")func DayOfWeek(year int, month int, day int) string {return time.Date(year, time.Month(month), day, 0, 0, 0, 0, time.Local).Weekday().String()}func main() {fmt.Println(DayOfWeek(2019, 8, 31))fmt.Println(DayOfWeek(1999, 7, 18))fmt.Println(DayOfWeek(1993, 8, 15))fmt.Println(DayOfWeek(1971, 6, 12))fmt.Println(DayOfWeek(2023, 2, 22))fmt.Println(DayOfWeek(2040, 6, 13))}
C++代码
ctime库
#include#includeusing namespace std;class Solution{public:string DayOfWeek(int year, int month, int day){const char *week[7] = {"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};struct tm t = {0};t.tm_year = year - 1900;t.tm_mon = month - 1;t.tm_mday = day;mktime(&t);return week[t.tm_wday%7];}};int main(){Solution s;cout << s.DayOfWeek(2019,8,31) << endl;cout << s.DayOfWeek(1999,7,18) << endl;cout << s.DayOfWeek(1993,8,15) << endl;cout << s.DayOfWeek(1971,6,12) << endl;cout << s.DayOfWeek(2023,2,22) << endl;cout << s.DayOfWeek(2040,6,13) << endl;return 0;}
输出：
Saturday
Sunday
Sunday
Saturday
Wednesday
Sunday
--------------------------------
Process exited after 0.02402 seconds with return value 0
请按任意键继续. . .
发现没？ 2040.6.13返回的星期是错的！
网上查了资料，原来ctime库的CTime对象是有指定范围的：
static CTime WINAPI GetCurrentTime( );
获取系统当前日期和时间。
返回表示当前日期和时间的CTime对象。
int GetYear( ) const；
获取CTime对象表示时间的年份。
范围从1970年1月1日到2038年1月18日。
时间范围测试：
#include#includeusing namespace std;class Solution{public:string DayOfWeek(int year, int month, int day){const char *week[7] = {"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};struct tm t = {0};t.tm_year = year - 1900;t.tm_mon = month - 1;t.tm_mday = day;mktime(&t);return week[t.tm_wday%7];}};int main(){Solution s;for (int i=16;i<25;i++)cout << i << ":" << s.DayOfWeek(2038,1,i) << endl;return 0;}
测试结果：
16:Saturday
17:Sunday
18:Monday
19:Tuesday
20:Sunday
21:Sunday
22:Sunday
23:Sunday
24:Sunday
--------------------------------
Process exited after 0.05159 seconds with return value 0
请按任意键继续. . .
2038.1.19日的星期也对，之后的全部返回Sunday。
修改这个问题，技术上一点问题都没有。目前C++都发展到C++20了，而我用的是C++11，暂不知道之后版本的库文件有没有对此问题作过更新。那么，问题来了：
之前用C语言写的的软件，用ctime或者time.h获取时间的软件在2038年1月19日之后都会发生错误。还好，还有整整15年时间来改正这个“时间Bug”。