如何获取给定月份的每个“星期一”？
一个例子;
输入：2017 年 7 月 11 日 (11.07.2017)
输出：( 3,10,17,24,31 )
2017年7月3日 星期一
2017年7月10日 星期一
2017年7月17日 星期一
2017年7月24日 星期一
2017年7月31日
我可以获得给定月份的天数（2017 年 7 月为 31 天）。然后编写一个迭代（for 循环 a.e.），如果 dayOfWeek 等于星期一，则添加到列表中。但这不是好代码，因为 for 循环将运行 31 次。应该有更好的算法来归档目标。
我正在使用 C# .net 框架 4.6
UPDATE
感谢大家的帮助，到目前为止我已经得到了一些答案；我用简单而肮脏的基准代码测试了所有代码，以找到更快的算法。
这是我的基准代码；
using System;
using System.Collections.Generic;
using System.Linq;
using BenchmarkDotNet.Attributes;
using BenchmarkDotNet.Attributes.Columns;
using BenchmarkDotNet.Attributes.Jobs;
using BenchmarkDotNet.Engines;
using X.Core.Helpers;
namespace X.ConsoleBenchmark
{
[SimpleJob(RunStrategy.ColdStart, targetCount: 5)]
[MinColumn, MaxColumn, MeanColumn, MedianColumn]
public class LoopTest
{
[Benchmark]
public void CalculateNextSalaryDateWithLoopAllDays()
{
DateTime date = new DateTime(2017, 7, 3);
const int oneMillion = 1000000;
for (int i = 0; i < oneMillion; i++)
{
List<DateTime> allXDaysInMonth = date.GetAllXDaysInMonthWithLoopAllDays(DayOfWeek.Tuesday);
if (allXDaysInMonth != null && allXDaysInMonth.FirstOrDefault().Day != 4)
{
throw new ApplicationException("Calculate method has errors.");
}
}
}
[Benchmark]
public void CalculateNextSalaryDate()
{
DateTime date = new DateTime(2017, 7, 3);
const int oneMillion = 1000000;
for (int i = 0; i < oneMillion; i++)
{
List<DateTime> allXDaysInMonth = date.GetAllXDaysInMonth(DayOfWeek.Tuesday);
if (allXDaysInMonth != null && allXDaysInMonth.FirstOrDefault().Day != 4)
{
throw new ApplicationException("Calculate method has errors.");
}
}
}
[Benchmark]
public void Maccettura_GetAllDayOfWeekPerMonth()
{
DateTime exampleDate = new DateTime(2017, 7, 3);
const int oneMillion = 1000000;
for (int i = 0; i < oneMillion; i++)
{
var date = new DateTime(exampleDate.Year, exampleDate.Month, 1);
if (date.DayOfWeek != DayOfWeek.Thursday)
{
int daysUntilDayOfWeek = ((int)DayOfWeek.Thursday - (int)date.DayOfWeek + 7) % 7;
date = date.AddDays(daysUntilDayOfWeek);
}
List<DateTime> days = new List<DateTime>();
while (date.Month == exampleDate.Month)
{
days.Add(date);
date = date.AddDays(7);
}
if (days.FirstOrDefault().Day != 6)
{
throw new ApplicationException("Calculate method has errors.");
}
}
}
[Benchmark]
public void ScottHannen_GetWeekdaysForMonth()
{
DateTime exampleDate = new DateTime(2017, 7, 3);
const int oneMillion = 1000000;
for (int i = 0; i < oneMillion; i++)
{
IEnumerable<DateTime> days = ScottHannen_GetDaysInMonth(exampleDate).Where(day => day.DayOfWeek == DayOfWeek.Thursday);
if (days.FirstOrDefault().Day != 6)
{
throw new ApplicationException("Calculate method has errors.");
}
}
}
private IEnumerable<DateTime> ScottHannen_GetDaysInMonth(DateTime date)
{
var dateLoop = new DateTime(date.Year, date.Month, 1);
while (dateLoop.Month == date.Month)
{
yield return dateLoop;
dateLoop = dateLoop.AddDays(1);
}
}
[Benchmark]
public void Trioj_GetWeekdaysForMonth()
{
DateTime exampleDate = new DateTime(2017, 7, 3);
const int oneMillion = 1000000;
for (int i = 0; i < oneMillion; i++)
{
IEnumerable<DateTime> days = Trioj_GetDatesInMonthByWeekday(exampleDate, DayOfWeek.Thursday);
if (days.FirstOrDefault().Day != 6)
{
throw new ApplicationException("Calculate method has errors.");
}
}
}
private List<DateTime> Trioj_GetDatesInMonthByWeekday(DateTime date, DayOfWeek dayOfWeek)
{
// We know the first of the month falls on, well, the first.
var first = new DateTime(date.Year, date.Month, 1);
int daysInMonth = DateTime.DaysInMonth(date.Year, date.Month);
// Find the first day of the week that matches the requested day of week.
if (first.DayOfWeek != dayOfWeek)
{
first = first.AddDays(((((int)dayOfWeek - (int)first.DayOfWeek) + 7) % 7));
}
// A weekday in a 31 day month will only occur five times if it is one of the first three weekdays.
// A weekday in a 30 day month will only occur five times if it is one of the first two weekdays.
// A weekday in February will only occur five times if it is the first weekday and it is a leap year.
// Incidentally, this means that if we subtract the day of the first occurrence of our weekday from the
// days in month, then if that results in an integer greater than 27, there will be 5 occurrences.
int maxOccurrences = (daysInMonth - first.Day) > 27 ? 5 : 4;
var list = new List<DateTime>(maxOccurrences);
for (int i = 0; i < maxOccurrences; i++)
{
list.Add(new DateTime(first.Year, first.Month, (first.Day + (7 * i))));
}
return list;
}
[Benchmark]
public void Jonathan_GetWeekdaysForMonth()
{
DateTime exampleDate = new DateTime(2017, 7, 3);
const int oneMillion = 1000000;
for (int i = 0; i < oneMillion; i++)
{
IEnumerable<DateTime> days = Jonathan_AllDatesInMonth(exampleDate.Year, exampleDate.Month).Where(x => x.DayOfWeek == DayOfWeek.Thursday);
if (days.FirstOrDefault().Day != 6)
{
throw new ApplicationException("Calculate method has errors.");
}
}
}
private static IEnumerable<DateTime> Jonathan_AllDatesInMonth(int year, int month)
{
int days = DateTime.DaysInMonth(year, month);
for (int day = 1; day <= days; day++)
{
yield return new DateTime(year, month, day);
}
}
[Benchmark]
public void Swatsonpicken_GetWeekdaysForMonth()
{
DateTime exampleDate = new DateTime(2017, 7, 3);
const int oneMillion = 1000000;
for (int i = 0; i < oneMillion; i++)
{
IEnumerable<DateTime> days = Swatsonpicken_GetDaysOfWeek(exampleDate, DayOfWeek.Thursday);
if (days.FirstOrDefault().Day != 6)
{
throw new ApplicationException("Calculate method has errors.");
}
}
}
private static IEnumerable<DateTime> Swatsonpicken_GetDaysOfWeek(DateTime startDate, DayOfWeek desiredDayOfWeek)
{
var daysOfWeek = new List<DateTime>();
var workingDate = new DateTime(startDate.Year, startDate.Month, 1);
var offset = ((int)desiredDayOfWeek - (int)workingDate.DayOfWeek + 7) % 7;
// Jump to the first desired day of week.
workingDate = workingDate.AddDays(offset);
do
{
daysOfWeek.Add(workingDate);
// Jump forward seven days to get the next desired day of week.
workingDate = workingDate.AddDays(7);
} while (workingDate.Month == startDate.Month);
return daysOfWeek;
}
[Benchmark]
public void AliaksandrHmyrak_GetWeekdaysForMonth()
{
DateTime exampleDate = new DateTime(2017, 7, 3);
const int oneMillion = 1000000;
for (int i = 0; i < oneMillion; i++)
{
IEnumerable<DateTime> days = AliaksandrHmyrak_GetDaysOfWeek(exampleDate, DayOfWeek.Thursday);
if (days.FirstOrDefault().Day != 6)
{
throw new ApplicationException("Calculate method has errors.");
}
}
}
private static List<DateTime> AliaksandrHmyrak_GetDaysOfWeek(DateTime date, DayOfWeek dayOfWeek)
{
var daysInMonth = DateTime.DaysInMonth(date.Year, date.Month);
var i = 1;
List<DateTime> result = new List<DateTime>(5);
do
{
var testDate = new DateTime(date.Year, date.Month, i);
if (testDate.DayOfWeek == dayOfWeek)
{
result.Add(testDate);
i += 7;
}
else
{
i++;
}
} while (i <= daysInMonth);
return result;
}
}
}
And this is the results table;
如果您愿意，我可以删除任何代码和图片名称
我标记了乔纳森的答案。简单、干净、更快（有趣）。
其他答案也有效，但我更喜欢使用 Jon Skeet 的 AllDaysInMonth 函数对于一个月中的每一天 https://stackoverflow.com/questions/9097027/foreach-day-in-month
public static IEnumerable<DateTime> AllDatesInMonth(int year, int month)
{
int days = DateTime.DaysInMonth(year, month);
for (int day = 1; day <= days; day++)
{
yield return new DateTime(year, month, day);
}
}
然后你可以像这样使用 LINQ 进行调用：
var mondays = AllDatesInMonth(2017, 7).Where(i => i.DayOfWeek == DayOfWeek.Monday);
但我想这取决于你要使用它多少次来决定是否值得将其分解为一个单独的函数。