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