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JavaSE 计算2个List集合中的交集、差集、并集、去重并集

时间:2020-08-01 00:13:37      阅读:116      评论:0      收藏:0      [点我收藏+]

标签:null   turn   for   class   main方法   nal   并集   vat   rgs   

VideoOrder.java

重写里面的equals和hashCode方法

class VideoOrder {
    private int price;
    private String title;

    public VideoOrder(String title, int price) {
        this.title = title;
        this.price = price;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o)
            return true;
        if (o == null || getClass() != o.getClass())
            return false;
        VideoOrder that = (VideoOrder) o;
        return price == that.price && Objects.equals(title, that.title);
    }

    @Override
    public int hashCode() {
        return Objects.hash(price, title);
    }

    public int getPrice() {
        return price;
    }

    public void setPrice(int price) {
        this.price = price;
    }

    public String getTitle() {
        return title;
    }

    public void setTitle(String title) {
        this.title = title;
    }

    @Override
    public String toString() {
        return "VideoOrder{" + "price=" + price + ", title=‘" + title + ‘\‘‘ + ‘}‘;
    }
}

main方法

    public static void main(String[] args) {
        List<VideoOrder> videoOrders1 = new ArrayList<>();
        videoOrders1.add(new VideoOrder("a课程", 22));
        videoOrders1.add(new VideoOrder("w课程", 200));
        videoOrders1.add(new VideoOrder("c课程", 100));
        videoOrders1.add(new VideoOrder("d课程", 33));
        videoOrders1.add(new VideoOrder("f课程", 1));
        List<VideoOrder> videoOrders2 = new ArrayList<>();
        videoOrders2.add(new VideoOrder("a课程", 22));
        videoOrders2.add(new VideoOrder("b课程", 18));
        videoOrders2.add(new VideoOrder("d课程", 33));
        videoOrders2.add(new VideoOrder("f课程", 1));
        videoOrders2.add(new VideoOrder("z课程", 22));
        // 交集,方式一
        // videoOrders1.retainAll(videoOrders2);
        // System.out.println(videoOrders1);
        //方式二
        List<VideoOrder> intersectionList = new ArrayList<>();
        for (VideoOrder videoOrder : videoOrders1) {
            if (videoOrders2.contains(videoOrder)) {
                intersectionList.add(videoOrder);
            }
        }
        System.out.println("交集=" + intersectionList);
        // videoOrders1差集,方式一
        // videoOrders1.removeAll(videoOrders2);
        // System.out.println(videoOrders1);
        //方式二
        List<VideoOrder> diffList1 = new ArrayList<>();
        for (VideoOrder videoOrder : videoOrders1) {
            if (!videoOrders2.contains(videoOrder)) {
                diffList1.add(videoOrder);
            }
        }
        System.out.println("订单1差集=" + diffList1);
        // videoOrders2差集
        // videoOrders2.removeAll(videoOrders1);
        // System.out.println(videoOrders2);
        List<VideoOrder> diffList2 = new ArrayList<>();
        for (VideoOrder videoOrder : videoOrders2) {
            if (!videoOrders1.contains(videoOrder)) {
                diffList2.add(videoOrder);
            }
        }
        System.out.println("订单2差集=" + diffList2);
        // 并集
         videoOrders1.addAll(videoOrders2);
         System.out.println(videoOrders1);
        // 去重并集
        // Set<VideoOrder> set = new HashSet<>(videoOrders1);
        // System.out.println(set);
    }

 

 

 

JavaSE 计算2个List集合中的交集、差集、并集、去重并集

标签:null   turn   for   class   main方法   nal   并集   vat   rgs   

原文地址:https://www.cnblogs.com/chenyanbin/p/13412669.html

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