Question

我有一个程序从db获取java对象的列表，并将它与已经检索的旧列表进行比较，并在其中找到delta（差异）元素并返回。我想知道是否有最好的方法来做这个，而不是只使用Set方法Union（），Intersection（）等，并避免内存不足错误？列表的大小可以是200k。我在我的项目中使用Spring 3.2.8.RELEASE版本。

public class Tester {

    private List<AddressInfo> oldListOfAddresses;

    @Scheduled(cron="0 1 6 * * ?") // 6 AM everyday
    public Map<String, AddressInfo> getCompany() {
        try {
            Map<String, AddressInfo> companyMap = new HashMap<>();
            String sql = "Some sql query which return Address Info.";
            List<AddressInfo> newListOfAddresses = jdbcTemplate.query(sql, new Object[0],
                    new FacilityNewMapper());
            if (newListOfAddresses == null || newListOfAddresses.size() = 0) {
                throw new FacilityLookUpException("List of clinic Info from facilities is empty...");
            } else {

                // I have to find the delta of new list and old list here.
                // I need an efficient (Space and Time) way of finding delta.
                List<AddressInfo> deltaList = newListOfAddresses - oldListOfAddresses; //Something like this

                for (AddressInfo comp : deltaList) {
                    if (comp != null) {
                        companyMap.put(comp.getLocationId(), comp);
                    }
                }
                oldListOfAddresses = newListOfAddresses;
            }
            return companyMap;
        } catch (Exception e) {
            throw new CompanyLookUpException(
                    "List of company addresses is empty..." + e.getMessage());
        }
    }
}

AddressInfo bean。

public class AddressInfo{

    private String locationId;
    private String streetName;
    private String city;
    private String state;
    private String country;

    public String getLocationId() {
        return locationId;
    }
    public void setLocationId(String locationId) {
        this.locationId = locationId;
    }
    public String getStreetName() {
        return streetName;
    }
    public void setStreetName(String streetName) {
        this.streetName = streetName;
    }
    public String getCity() {
        return city;
    }
    public void setCity(String city) {
        this.city = city;
    }
    public String getState() {
        return state;
    }
    public void setState(String state) {
        this.state = state;
    }
    public String getCountry() {
        return country;
    }
    public void setCountry(String country) {
        this.country = country;
    }
    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((city == null) ? 0 : city.hashCode());
        result = prime * result + ((country == null) ? 0 : country.hashCode());
        result = prime * result + ((locationId == null) ? 0 : locationId.hashCode());
        result = prime * result + ((state == null) ? 0 : state.hashCode());
        result = prime * result + ((streetName == null) ? 0 : streetName.hashCode());
        return result;
    }
    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        AddressInfo other = (AddressInfo) obj;
        if (city == null) {
            if (other.city != null)
                return false;
        } else if (!city.equals(other.city))
            return false;
        if (country == null) {
            if (other.country != null)
                return false;
        } else if (!country.equals(other.country))
            return false;
        if (locationId == null) {
            if (other.locationId != null)
                return false;
        } else if (!locationId.equals(other.locationId))
            return false;
        if (state == null) {
            if (other.state != null)
                return false;
        } else if (!state.equals(other.state))
            return false;
        if (streetName == null) {
            if (other.streetName != null)
                return false;
        } else if (!streetName.equals(other.streetName))
            return false;
        return true;
    }

}

Answer 1

我不这么认为（注意：我认为列表的顺序并不重要）。例如，在不使用set的情况下执行此操作的最快方法是对两个列表进行排序，这些列表将花费您O（nlogn），然后迭代它们比较每个元素并保存那些没有对的列表。在Set的情况下，您将基本迭代每个元素并在第二组中查找它，因此迭代为O（n）并且搜索为O（1）。最后我们有O（nlogn）＆gt; O（n）集胜

Answer 2

假设AddressInfo正确实施equals和hashCode，并且每个列表中的项目都是唯一的，则以下函数可以在线性时间内找到增量：

Set< AddressInfo > findDiff(final List< AddressInfo > newListOfAddresses, final List< AddressInfo > oldListOfAddresses) {
    Map< AddressInfo, Boolean > map = new HashMap<>(newListOfAddresses.size());

    for ( AddressInfo addressInfo : newListOfAddresses ) {
        map.put( addressInfo, TRUE );
    }

    for ( AddressInfo addressInfo : oldListOfAddresses ) {
        map.remove( addressInfo );
    }

    return map.keySet();
}

Answer 3

这应该可以很好地创建两个列表之间的差异。

这里我创建一个集合并添加newList的所有元素。然后，无论哪个元素都是oldList的一部分，我将删除它们。

Set< AddressInfo > findDiffOfTwoList( List< AddressInfo > newList, List< AddressInfo > oldList) {
    Set<AddressInfo> set = new HashSet<>();
    set.addAll(newList);
    for(AddressInfo address:oldList){
        set.remove(address);
    }
    return set;
}

在Java中比较两个列表的有效方法是什么？

3 个答案: