输入描述:输入包含一个八位整数 N,表示日期。对于所有评测用例,10000101 ≤ N ≤ 89991231,保证 N 是一个合法日期的 8 位数
输出描述:输出两行,每行 1 个八位数。第一行表示下一个回文日期,第二行表示下一个 ABABBABA 型回文日期
一.首先创建一个包和类
- 创建静态方法isLegalDate(String strDate)
四.完整代码
package t01;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Scanner;
public class PalindromicDate {
public static void main(String[] args) {
String strDate;
int year, month, day;
Scanner sc = new Scanner(System.in);
SimpleDateFormat sdf = new SimpleDateFormat("yyyydddd");
System.out.print("输入8位数构成的日期:");
strDate = sc.next();
if (isLegalDate(strDate)) {
System.out.println("[" + strDate + "]是合法日期");
//将用户输入的合法日期作为日期循环的起点
year = Integer.parseInt(strDate.substring(0, 4));
month = Integer.parseInt(strDate.substring(4, 6));
day = Integer.parseInt(strDate.substring(6));
Calendar calendar = Calendar.getInstance();
calendar.set(year, month - 1, day);
//任务1 输出该日期之后的第一个回文日期
String strDate1 = strDate;
Calendar calender1 = calendar;
while (true) {
calender1.add(Calendar.DAY_OF_MONTH, 1);//往后推1天
strDate1 = sdf.format(calender1.getTime());
if (isLegalDate(strDate1)) {
break;//找到一个回文日期就跳出循环
}
}
System.out.println("[" + strDate1 + "]是回文日期");
} else {
System.out.println("[" + strDate + "]是非法日期");
}
}
private static boolean isLegalDate(String strDate) {
int year, month, day;
year = Integer.parseInt(strDate.substring(0, 4));
month = Integer.parseInt(strDate.substring(4, 6));
day = Integer.parseInt(strDate.substring(6));
//利用反向思维来处理
if (year < 1000 || year > 8999) return false;
if (month < 1 || month > 12) return false;
if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12) {
if (day < 1 || day > 31) return false;
} else if (month == 2) {
if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {//闰年判断
if (day < 1 || day > 29) return false;
} else {
if (day < 1 || day > 28) return false;
}
} else {
if (day < 1 || day > 30) return false;
}
return true;
}
private static boolean isPalindromicDate(String strDate) {
for (int i = 0; i < 4; i++) {
//采用反向思维
if (strDate.charAt(i) != strDate.charAt(7 - i)) ;
}
return true;
}}
package t01;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Scanner;
public class PalindromicDate {
public static void main(String[] args) {
String strDate;
int year, month, day;
Scanner sc = new Scanner(System.in);
SimpleDateFormat sdf = new SimpleDateFormat("yyyydddd");
System.out.print("输入8位数构成的日期:");
strDate = sc.next();
if (isLegalDate(strDate)) {
System.out.println("[" + strDate + "]是合法日期");
//将用户输入的合法日期作为日期循环的起点
year = Integer.parseInt(strDate.substring(0, 4));
month = Integer.parseInt(strDate.substring(4, 6));
day = Integer.parseInt(strDate.substring(6));
Calendar calendar = Calendar.getInstance();
calendar.set(year, month - 1, day);
//任务1 输出该日期之后的第一个回文日期
String strDate1 = strDate;
Calendar calender1 = calendar;
while (true) {
calender1.add(Calendar.DAY_OF_MONTH, 1);//往后推1天
strDate1 = sdf.format(calender1.getTime());
if (isLegalDate(strDate1)) {
break;//找到一个回文日期就跳出循环
}
}
System.out.println("[" + strDate1 + "]是回文日期");
} else {
System.out.println("[" + strDate + "]是非法日期");
}
}
private static boolean isLegalDate(String strDate) {
int year, month, day;
year = Integer.parseInt(strDate.substring(0, 4));
month = Integer.parseInt(strDate.substring(4, 6));
day = Integer.parseInt(strDate.substring(6));
//利用反向思维来处理
if (year < 1000 || year > 8999) return false;
if (month < 1 || month > 12) return false;
if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12) {
if (day < 1 || day > 31) return false;
} else if (month == 2) {
if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {//闰年判断
if (day < 1 || day > 29) return false;
} else {
if (day < 1 || day > 28) return false;
}
} else {
if (day < 1 || day > 30) return false;
}
return true;
}
private static boolean isPalindromicDate(String strDate) {
for (int i = 0; i < 4; i++) {
//采用反向思维
if (strDate.charAt(i) != strDate.charAt(7 - i)) ;
}
return true;
}}



