一、了解两个时间比大小

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;

public class TimeCompare{
    public static void main(String[] args) {
        String time1 = "2022-01-01 12:00:00";
        String time2 = "20212-01-01 13:00:00";

        SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

        try {
            Date date1 = sdf.parse(time1);
            Date date2 = sdf.parse(time2);

            if (date1.before(date2)) {
                System.out.println(time1 + " 在 " + time2 + " 之前");
            } else if (date1.after(date2)) {
                System.out.println(time1 + " 在 " + time2 + " 之后");
            } else {
                System.out.println(time1 + " 和 " + time2 + " 相等");
            }
        } catch (ParseException e) {
            e.printStackTrace();
        }
    }
}

二、自己的需求

public class TimeOverlap {  
    //时间一和时间二是否重叠
     public static boolean isOverLap(String start1, String end1, String start2, String end2) {
        try {
            SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
            Date dateTime1Start = sdf.parse(start1);
            Date dateTime1End = sdf.parse(end1);
            Date dateTime2Start = sdf.parse(start2);
            Date dateTime2End = sdf.parse(end2);
            // 检查时间段1的开始时间是否在时间段2的结束时间之后，或者时间段1的结束时间是否在时间段2的开始时间之前
            return !(dateTime1End.before(dateTime2Start) || dateTime1Start.after(dateTime2End));
        } catch (ParseException e) {
//            throw new RuntimeException(e);
            // 处理日期时间解析异常
            System.err.println("无法解析日期时间: " + e.getMessage());
            return false;
        }
    }
   //时间一包含时间二
    public static boolean isContainLap(String start1, String end1, String start2, String end2) {
        try {
            SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
            Date dateTime1Start = sdf.parse(start1);
            Date dateTime1End = sdf.parse(end1);
            Date dateTime2Start = sdf.parse(start2);
            Date dateTime2End = sdf.parse(end2);
            return !(dateTime1Start.before(dateTime2Start) && dateTime1End.after(dateTime2End));
        } catch (ParseException e) {
            // 处理日期时间解析异常
            System.err.println("无法解析日期时间: " + e.getMessage());
            return false;
        }
    }
//时间二包含时间一
    public static boolean isContainNextLap(String start1, String end1, String start2, String end2) {
        try {
            SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
            Date dateTime1Start = sdf.parse(start1);
            Date dateTime1End = sdf.parse(end1);
            Date dateTime2Start = sdf.parse(start2);
            Date dateTime2End = sdf.parse(end2);
            return !(dateTime2Start.before(dateTime1Start) && dateTime2End.after(dateTime1End));
        } catch (ParseException e) {
            // 处理日期时间解析异常
            System.err.println("无法解析日期时间: " + e.getMessage());
            return false;
        }
    }
//判断时间一开始时间是否大于时间二开始时间
    public static boolean isContainLtLap(String start1, String end1, String start2, String end2) {
        try {
            SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
            Date dateTime1Start = sdf.parse(start1);
            Date dateTime1End = sdf.parse(end1);
            Date dateTime2Start = sdf.parse(start2);
            Date dateTime2End = sdf.parse(end2);
            return !(dateTime1Start.before(dateTime2Start) && dateTime1End.before(dateTime2End));
        } catch (ParseException e) {
            // 处理日期时间解析异常
            System.err.println("无法解析日期时间: " + e.getMessage());
            return false;
        }
    }
    public static void main(String[] args) {  
        String start1 = "2023-08-24 09:00:00";  
        String end1 = "2023-08-24 10:00:00";  
  
        String start2 = "2023-08-24 09:30:00";  
        String end2 = "2023-08-24 11:00:00";  
  
        System.out.println(isOverlap(start1, end1, start2, end2)); // 输出：true  
    }  
}

三、进行时间比对

//如果返回true，则说明时间有交集
if (isOverLap(start1, end1, start2, end2)){
    //判断时间1是否包含时间二
   if(isContainLap(start1, end1, start2, end2)){
    //
    }else if(isContainNextLap(start1, end1, start2, end2)){
    //判断时间二是否包含时间一
    }else if(isContainLtLap(start1, end1, start2, end2)){
      //判断时间一开始时间是否大于时间二开始时间
    }else{
    }

}