Regular Expression with leap year judgment

Author: Xixigongzhu (Princess Xi)Http://search.csdn.net/Expert/topic/1974/1974227.xml? Temp =. 6640131.

First, your year range is 1800-3999
Runnian:
1800,1900, 2100,2200, 2300,2500, 2600,2700, 2900,3000, 3100,3300, 3400,3500, 3700,3800, 3900 is a special value (which can be divisible by four but not a runner-up:
String leap1 = "(1 [8-9]) | ([2-3] [0-9]) (0 | 2 | 4 | 6 | 8) (4 | 8 ))";
String leap2 = "(2 (0 | 4 | 8) | (3 (2 | 6) 00 )";
Other 0 ends:
String leap3 = "(1 [8-9]) | ([2-3] [0-9]) (2 | 4 | 6 | 8) 0) ";
Ended with 2, 6:
String leader = "(1 [8-9]) | ([2-3] [0-9]) (1 | 3 | 5 | 7 | 9) (2 | 6 ))";
February:
String leapmonth = "(02-([0-1] [1-9]) | (10) | (2 [0-8])";
Other months:
String other1 = "(01) | (0 [3-9]) | (1 [0-2]) -([0-2] [1-9]) | ([1-3] 0 )))";
String other2 = "(0 (1 | 3 | 5 | 7 | 8) | (10) | (12)-(31 ))";
Use | to connect the above four regular expressions, and then add the processing of the first month of the last year and the other months:
String leap = "(" + leap1 + "|" + leap2 + "|" + leap3 + "|" + leap + ")-("
+ Leapmonth + "|" + other1 + "|" + other2 + ")";

Non-Profit Year:
String noleap1 = "(1 [8-9]) | ([2-3] [0-9]) (0 | 2 | 4 | 6 | 8) (1 | 2 | 3 | 5 | 6 | 7 | 9 ))";
String noleap2 = "(1 [8-9]) | (2 (1 | 2 | 3 | 5 | 6 | 7 | 9 )) | (3 (0 | 1 | 3 | 4 | 5 | 7 | 8 | 9) 00 )";
String noleap3 = "(1 [8-9]) | ([2-3] [0-9]) (1 | 3 | 5 | 7 | 9) (0 | 1 | 3 | 4 | 5 | 7 | 8 | 9 ))";
Non-profit year February:
String month = "(02-([0-2] [1-9]) | ([1-2] 0 )))";
Use | to connect the above three regular expressions, and then add the processing of the non-runner-up year on January 1, February and other months:
String noleap = "(" + noleap1 + "|" + noleap2 + "|" + noleap3 + ")-("
+ Month + "|" + other1 + "|" + other2 + ")";

The combination of runnian and non-runnian regular expressions is:
String yearregex = leap + "|" + noleap;