java8 stream取出 最大值/最小值

注:转载请注明出处!!!

这里直接用取出多个对象中某个值 最大/最小 来进行举例

直接看代码

 /**
  * 时间测试类
  */
 class TimeTest {
     private Date time;

     public Date getTime() {
         return time;
     }

     public void setTime(Date time) {
         this.time = time;
     }
 }

工具类:

 public class DateUtils extends org.apache.commons.lang.time.DateUtils{
     //自定义日期格式
     public static final String SIMPLE_DATEFORMAT = "yyyy-MM-dd HH:mm:ss";

     public static final String SIMPLE_DATEFORMAT_YMD = "yyyy-MM-dd";
     //日期比较方法
     public static int compareDate(Date first, Date second){
        return first.compareTo(second);
     }
     //日期转string
    public static String formatDateYMDHMS(Date date){
         if(date==null){
             return null;
         }
         return format(date,SIMPLE_DATEFORMAT);
     }
     //通用格式化
     public static String format(Date date, String pattern) {
         if (date == null) {
             return null;
         }
         SimpleDateFormat format = new SimpleDateFormat(pattern);
         return format.format(date);
     }
 }

运行:

 public static void main(String[] args) {
         int i = 0;
         List<TimeTest> timeList = new ArrayList<>();
         while (++i <= 5) {
             TimeTest time = new TimeTest();
             time.setTime(DateUtils.addMonths(DateUtils.now(), i));
             timeList.add(time);
         }
         //插入一个空值
         TimeTest time = new TimeTest();
         timeList.add(time);
         System.out.println("---------------初始值------------------");
         timeList.forEach(o -> {
             System.out.println(DateUtils.formatDateYMDHMS(o.getTime()));
         });
         System.out.println("---------------------------------------");
         //下面有两种写法 推荐使用方法一
         //代码解释: 将timeList流首先过滤时间不为空的,将需要比较的值转出map然后去重,最后取出最大值/最小值
         //方法一 min(DateUtils::compareDate)  这种写法需要自己写工具类,且compareDate必须是static修饰
         Date min = timeList.stream().filter(o -> o.getTime() != null).map(TimeTest::getTime).distinct().min(DateUtils::compareDate).get();
         //方法二 max((e1, e2) -> e1.compareTo(e2))这种写法需要类里面含有比较方法
         Date max = timeList.stream().filter(o -> o.getTime() != null).map(TimeTest::getTime).distinct().max((e1, e2) -> e1.compareTo(e2)).get();
         System.out.println("min=" + DateUtils.formatDateYMDHMS(min));
         System.out.println("max=" + DateUtils.formatDateYMDHMS(max));
     }

输出:

 ---------------初始值------------------
 2019-09-09 14:35:53
 2019-10-09 14:35:53
 2019-11-09 14:35:53
 2019-12-09 14:35:53
 2020-01-09 14:35:53
 null
 ---------------------------------------
 min=2019-09-09 14:35:53
 max=2020-01-09 14:35:53

 Process finished with exit code 0