Question

以下课程的输出是：大小是3 大小是1

但是，如果我将TreeSet更改为HashSet，那么行：

Set<SuggestionDetailBean> set = new TreeSet<SuggestionDetailBean>();

变为

Set<SuggestionDetailBean> set = new HashSet<SuggestionDetailBean>();

输出是：大小是3 大小是2

使用HashSet或TreeSet喊出不改变Set的大小？使用HashSet似乎表现得像预期的那样，因为它正在删除重复项，但是当我使用TreeSet时，重复项仍然存在？我认为SuggestionDetailBean中的hashcode和equals方法是否被正确覆盖？

以下是代码：

public class TestSet {

    public static void main(String args[]){

        SuggestionDetailBean s = new SuggestionDetailBean();
        s.setTagList("teddst");
        s.setUrl("testurl");

        SuggestionDetailBean s2 = new SuggestionDetailBean();
        s2.setTagList("teddst");
        s2.setUrl("testurl");

        SuggestionDetailBean s3 = new SuggestionDetailBean();
        s3.setTagList("tessdafat");
        s3.setUrl("fdfaasdfredtestur ldd");

        List<SuggestionDetailBean> list = new ArrayList<SuggestionDetailBean>();
        list.add(s);
        list.add(s2);
        list.add(s3);

        Set<SuggestionDetailBean> set = new TreeSet<SuggestionDetailBean>();
        set.addAll(list);

        System.out.println("size is "+list.size());
        System.out.println("size is "+set.size());

    }

}

public class SuggestionDetailBean implements Comparable<Object> {

    private String url;
    private String tagList;
    private String numberOfRecommendations;
    private String date;
    private String time;
    private String summary;
    private String truncatedUrl;


    public void setTruncatedUrl(String truncatedUrl) {

        if(truncatedUrl.length() > 20){
            truncatedUrl = truncatedUrl.substring(0, 20)+"...";
        }

        this.truncatedUrl = truncatedUrl;
    }

    public String getSummary() {
        if(summary == null){
            return "";
        }
        else {
            return summary;
        }
    }

    public void setSummary(String summary) {
        this.summary = summary;
    }

    public String getDate() {
        return date;
    }

    public void setDate(String date) {
        this.date = date;
    }


        public String getTime() {
            return time;
        }

        public String getTruncatedUrl() {
            return this.truncatedUrl;
        }

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

        public String getTagList() {
            if(tagList == null){
                return "";
            }
            else {
                return tagList;
            }
        }

        public void setTagList(String tagList) {
            this.tagList = tagList;
        }


        public String getUrl() {
            return url;
        }

        public void setUrl(String url) {
            this.url = url;
        }

        public String getNumberOfRecommendations() {
            return numberOfRecommendations;
        }

        public void setNumberOfRecommendations(String numberOfRecommendations) {
            this.numberOfRecommendations = numberOfRecommendations;
        }

        @Override
        public int compareTo(Object o) {

            DateFormat formatter;
            Date date1 = null;
            Date date2 = null;  
            SuggestionDetailBean other = (SuggestionDetailBean) o;

            if(this.date == null || other.date == null){
                return 0;
            }   
            formatter = new SimpleDateFormat(SimpleDateFormatEnum.DATE.getSdfType()+" "+SimpleDateFormatEnum.TIME.getSdfType());
            try {
                date1 = (Date) formatter.parse(this.date + " " + this.time);
                date2 = (Date) formatter.parse(other.date + " " + other.time);
            } catch (ParseException e) {
                System.out.println("Exception thrown in"+this.getClass().getName()+", compareTo method");
                e.printStackTrace();
            }
            catch(NullPointerException npe){
                System.out.println("Exception thrown "+npe.getMessage()+" date1 is "+date1+" date2 is "+date2);
            }

             return date2.compareTo(date1);

        }

        @Override
           public int hashCode() {
                return this.url.hashCode();
            }

        @Override
        public boolean equals(Object obj) {

            SuggestionDetailBean suggestionDetailBean = (SuggestionDetailBean) obj;

            if(StringUtils.isEmpty(this.getTagList())){
                return this.getUrl().equals(suggestionDetailBean.getUrl());
            }
            else {
                return (this.getTagList().equals(suggestionDetailBean.getTagList())) &&
                        (this.getUrl().equals(suggestionDetailBean.getUrl()));
            }

        }

    }

编辑：注意：如果我使用以下方法将hashset转换为树集：

 Set<SuggestionDetailBean> sortedSet = new TreeSet<SuggestionDetailBean>(hashset);

然后保持正确的排序，因为删除重复项基于对象哈希码和equals方法而不是compareto方法。

Answer 1

根据the Javadoc for TreeSet：

注意由一组维护的排序（无论是否显式提供比较器）如果是，则必须与equals 一致正确实现Set接口。（见Comparable 或Comparator以获得与一致的精确定义等于。）这是因为Set接口定义在 equals操作的术语，但是TreeSet实例使用compareTo执行所有元素比较（或 compare）方法，因此这个方法认为两个元素相等从集合的角度来看，是平等的。一组的行为即使它的排序与equals不一致，也是明确定义的;它只是没有遵守Set接口的一般合同。

所以，问题出在你的compareTo方法上：要么是提供不一致的结果，要么就是提供一致的结果，这些结果不符合a.compareTo(b) == 0的规则，当且仅当{{1} }。

例如，这一位：

a.equals(b)

表示“如果if(this.date == null || other.date == null){ return 0; }或this有other，则报告date == null和this相等”，这肯定不是您想要的

HashSet删除重复项但TreeSet没有？

1 个答案: