我的用户对象如下所示：
User.java：
public class User {
public String firstName;
public String lastName;
public String getFirstName() {
return firstName;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public String getLastName() {
return lastName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
@Override
public int hashCode() {
return (this.firstName.hashCode() + this.lastName.hashCode());
}
@Override
public boolean equals(Object obj) {
if(obj instanceof User) {
User temp = (User) obj;
if(this.firstName.equals(temp.firstName) && this.lastName.equals(temp.lastName)) {
return true;
}
}
return false;
}
}
主程序如下所示：
import java.util.*;
class pp {
public static void main(String[] args) {
List<User[]> a = new ArrayList<User[]>();
User[] u = new User[3];
u[0] = new User();
u[0].setFirstName("Mike"); u[0].setLastName("Jordon");
u[1] = new User();
u[1].setFirstName("Jack"); u[1].setLastName("Nicolson");
u[2] = new User();
u[2].setFirstName("Jack"); u[2].setLastName("Nicolson");
a.add(u);
Set<User[]> s = new HashSet<User[]>(a);
for (User[] ss : s) {
for (int i=0; i<ss.length; i++) {
System.out.println(ss[i].getFirstName() + " " + ss[i].getLastName());
}
}
}
}
我期待输出
Mike Jordon
Jack Nicolson
但不知何故，它保留了重复对象＆amp;印刷为：
Mike Jordon
Jack Nicolson
Jack Nicolson
任何人都可以告诉我我错过了什么吗？
谢谢！
答案 0 :(得分：6)
你的equals方法应该是：
@Override
public boolean equals(Object obj) {
if(obj instanceof User) {
User temp = (User) obj;
if(this.firstName.equals(temp.firstName) && this.lastName.equals(temp.lastName)) {
return true;
}
}
return false;
}
答案 1 :(得分：5)
我已经完成了你的问题并理解了这个要求。请查找我已实现的类似代码，并成功从具有重复值的集合中删除对象。
@Snipet...
Employee.java
==============
package com.hcl;
public class Employee {
public String empid;
public String empname;
public double sal;
public int age;
public Employee(){
}
public Employee(String empid,String empname,double sal,int age){
this.empid = empid;
this.empname = empname;
this.sal = sal;
this.age = age;
}
public String getEmpid() {
return empid;
}
public void setEmpid(String empid) {
this.empid = empid;
}
public String getEmpname() {
return empname;
}
public void setEmpname(String empname) {
this.empname = empname;
}
public double getSal() {
return sal;
}
public void setSal(double sal) {
this.sal = sal;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
/**
* This override method playes a major role to remove duplicate values
*/
@Override
public int hashCode() {
return (this.empid.hashCode() + this.empname.hashCode()+String.valueOf(this.sal).hashCode()+String.valueOf(this.age).hashCode());
}
/**
* This override method plays a major role to remove duplicate values
*/
@Override
public boolean equals(Object obj) {
if(obj instanceof Employee) {
Employee temp = (Employee) obj;
if(this.empid.equals(temp.empid) && this.empname.equals(temp.empname) && String.valueOf(this.sal).equals(String.valueOf(temp.sal)) && String.valueOf(this.age).equals(String.valueOf(temp.age))) {
return true;
}
}
return false;
}
}
@Snipet..........
RemoveDuplicateObjects.java
=============================
package com.hcl;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class RemoveDuplicateObjects {
public static void main(String[] args) {
Employee emp1 = new Employee("1","bapi",1000,31);
Employee emp2 = new Employee("2","mano",2000,29);
Employee emp3 = new Employee("1","bapi",1000,31); // emp3 == emp1 duplicate object
Employee emp4 = new Employee("3","Rohan",3000,27);
Employee emp5 = new Employee("1","bapi",1000,31); // emp5 == emp3 == emp1 duplicate object
RemoveDuplicateObjects obj = new RemoveDuplicateObjects();
// empList contains objects having duplicate values. How to remove duplicate?
List<Employee> empList = new ArrayList<Employee>();
empList.add(emp1);
empList.add(emp2);
empList.add(emp3);
empList.add(emp4);
empList.add(emp5);
if(emp1.equals(emp2)){
System.out.println("emp1 and emp2 are equal");
}
if(emp1.equals(emp3)){
System.out.println("emp1 and emp3 are equal");
}
obj.removeDuplicate(empList);
}
// method is used for removing objects having duplicate values
private void removeDuplicate(List<Employee> empList) {
Set<Employee> empSet = new HashSet<Employee>();
empSet.addAll(empList);
for(Employee e: empSet){
System.out.println("id = "+e.getEmpid());
System.out.println("name = "+e.getEmpname());
System.out.println("sal = "+e.getSal());
System.out.println("age = "+e.getAge());
}
}
}
Done! Now you can run the program and analyze the solution.
答案 2 :(得分：1)
试试我的朋友：
Iterator i = a.iterator();
while (i.hasNext()) {
User u = (User) i.next();
boolean match = false;
Iterator j = a.iterator();
boolean once = true;
while (j.hasNext()) {
if(once){j.next();} // to skip own occurence only once
once = false;
User u2 = (User) j.next();
if (u.getFirstName().equals(u2.getFirstName())
&& u.getLastName().equals(u2.getLastName())) {
match = true;
}
}
if (!match) {
// print
}
}
答案 3 :(得分：0)
通常，您可以向Set添加元素以删除重复项。但是，您通常不希望将整个数组添加到集合中;你只想添加各个元素，如下所示：
public static void main(String[] args) {
Set<User> a = new HashSet<User>();
User[] u = new User[3];
u[0] = new User();
u[0].setFirstName("Mike"); u[0].setLastName("Jordon");
u[1] = new User();
u[1].setFirstName("Jack"); u[1].setLastName("Nicolson");
u[2] = new User();
u[2].setFirstName("Jack"); u[2].setLastName("Nicolson");
// Add each of the users to the Set. Note that there are three.
for (User user : u) {
a.add(u);
}
// Get the results back as an array. Note that this will have two.
User[] duplicatesRemoved = new User[0];
a.toArray(duplicatesRemoved);
}
答案 4 :(得分：0)
我想你想要这个：
class pp {
public static void main(String[] args) {
Set<User> a = new HashSet<User>();
User u = new User();
u.setFirstName("Mike"); u.setLastName("Jordon");
a.add(u);
u = new User();
u.setFirstName("Jack"); u.setLastName("Nicolson");
a.add(u);
u = new User();
u.setFirstName("Jack"); u.setLastName("Nicolson");
a.add(u);
for (User ss : a) {
System.out.println(ss.getFirstName() + " " + ss.getLastName());
}
}
}
答案 5 :(得分：0)
按照Jason的建议覆盖equals方法。 现在要删除重复项，您需要使用Set。
列表允许重复值，因此您将始终具有重复值。 Set不允许重复值，因此它可以解决您的问题。
答案 6 :(得分：0)
您正在使用一组数组，其中该集合包含一个元素，即一个包含三个用户的数组。数组不强制执行或检查唯一性，这就是为什么两次获得相同用户的原因。如果您完全删除了数组，并且只使用了一个Set，那么您将获得所需的“独特”行为。
答案 7 :(得分：0)
首先，如果你不想复制，你应该使用Set来存储对象而不是数组。 （数组和列表允许重复的对象存储）
其次你的equals方法应该使用String.equal方法进行比较，并且应该检查空值是否安全。我会将IDE的自动生成功能用于hashcode，并始终使用equals方法（即Eclipse Source - ＆gt; Generate hashCode（）和equals（）...）
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((firstName == null) ? 0 : firstName.hashCode());
result = prime * result + ((lastName == null) ? 0 : lastName.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;
User other = (User) obj;
if (firstName == null) {
if (other.firstName != null)
return false;
} else if (!firstName.equals(other.firstName))
return false;
if (lastName == null) {
if (other.lastName != null)
return false;
} else if (!lastName.equals(other.lastName))
return false;
return true;
}
和主要方法
public static void main(String[] args) {
List<Set<User>> a = new ArrayList<Set<User>>();
Set<User> set = new HashSet<User>();
User u = new User();
u.setFirstName("Mike"); u.setLastName("Jordon");
set.add(u);
u = new User();
u.setFirstName("Jack"); u.setLastName("Nicolson");
set.add(u);
u = new User();
u.setFirstName("Jack"); u.setLastName("Nicolson");
set.add(u);
a.add(set);
for (Set<User> ss : a) {
for (User user : ss) {
System.out.println(user.getFirstName() + " " + user.getLastName());
}
}
}
答案 8 :(得分：0)
您好，您可以在pp类中编写一个方法，以便从用户数组中删除重复的元素，如下所示：
private User[] getUserArrayWithoutDuplicates(User[] a) {
int count = a.length;
Set<User> tempset = new HashSet<User>();
for (int i = 0; i < count; i++) {
User[] user = a;
int arraysize = user.length;
for (int j = 0; j < arraysize; j++)
tempset.add(user[j]);
}
User[] usr = new User[tempset.size()];
Iterator<User> tempIterator = tempset.iterator();
int p = 0;
while (tempIterator.hasNext()) {
User user = tempIterator.next();
usr[p] = new User();
usr[p].setFirstName(user.firstName);
usr[p].setLastName(user.lastName);
p++;
}
return usr;
}
此方法将从User数组中删除重复的条目，并返回没有重复条目的User数组。