你可以这样做。查看java.time包以获得更多有用的日期/时间类。

import java.time.LocalDate;
import java.time.format.DateTimeFormatter;
import java.time.temporal.ChronoUnit;
LocalDate start = LocalDate.of(1910,1,1);
LocalDate end = LocalDate.of(2025,12,31);
DateTimeFormatter fmt = DateTimeFormatter.ofPattern("MMMM d, yyyy");
int key = 1;
while (start.isBefore(end)) {
    System.out.println(key++ + " : " + start.format(fmt));
    start = start.plus(1, ChronoUnit.DAYS);
}

像这样的指纹。

1 : January 1, 1910
2 : January 2, 1910
3 : January 3, 1910
4 : January 4, 1910
5 : January 5, 1910
6 : January 6, 1910
7 : January 7, 1910
8 : January 8, 1910
9 : January 9, 1910
10 : January 10, 1910
11 : January 11, 1910
12 : January 12, 1910
...
...

另一方面，如果不想使用任何导入的类来执行此操作，则可以按以下方式执行：
创建月份名称数组
创建每月最大天数（非闰年）的数组。
创建闰年方法（稍后解释）。

String[] monthNames = { "January", "February", "March",
        "April", "May", "June", "July", "August", "September",
        "October", "November", "December" };
int[] monthDays =
        { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int day = 1;
int month = 0; // months go from 0 to 11
int year = 1910;
int count = 0;

一个简单的while循环驱动程序来打印日期范围。

while (!(year == 2025 && month == 11 && day == 31)) {
    count++;
    System.out.printf("%d : %s %d, %d%n", count,
            monthNames[month], day, year);
     if (day++ >= monthDays[month]) {
         // take care of February in a leap year
         if (month == 1 && day == 29 && isLeap(year)) {
             continue;
         }
         day = 1;        // reset day if past end of month.
         month++;
         if (month == 12) {
             month = 0;  // reset month and increment year
             year++;
         } 
     }
}

只有能被4整除的非世纪年和能被400整除的世纪年才是闰年。首先检查非世纪年，因为它们更频繁地出现。

public static boolean isLeap(int year) {
    return year % 100 != 0 && year % 4 == 0 || year % 400 == 0;
}

这是一种生成所请求的数据的方法，正确地（人们希望）计算闰年等。我将注意到，运行的最后一个数字与示例输出的数字不同。
如果需要将数据存储到某个数据数组中，请替换 output 方法。

private static DateTimeFormatter FMTR = DateTimeFormatter.ofPattern("MMMM d, yyyy");
    public static LocalDate genNextDate(LocalDate inpDate)
    {
        return inpDate.plusDays(1);
    }
    public static void output(int cntr, LocalDate inpDate)
    {
        System.out.printf("%6d: %s%n", cntr, FMTR.format(inpDate));
    }
    public static void doIt()
    {
        final LocalDate endDate = LocalDate.of(2025, 12, 31);
        int cntr = 1;
        LocalDate date = LocalDate.of(1910, 1, 1);
        while (date.isBefore(endDate)) {
            output(cntr, date);
            date = genNextDate(date);
            cntr++;
        }
    }
   public static void main(String[] args)
    {
        doIt();
    }

示例输出

1: January 1, 1910
     2: January 2, 1910
     3: January 3, 1910
     4: January 4, 1910
     5: January 5, 1910
     6: January 6, 1910
     7: January 7, 1910
     8: January 8, 1910
     9: January 9, 1910
     ...
 42364: December 26, 2025
 42365: December 27, 2025
 42366: December 28, 2025
 42367: December 29, 2025
 42368: December 30, 2025