一般而言，数组最简单快捷的方法是直接利用HashSet的不重复的特性就可以做到，或者是遍历时利用List的contains 判断是否存在就可以去重了
如果是对象数组，对象的类需要 重写 equal 和 hashCode方法.。
但是面试题经常有数组去重的试题，如果在不使用hashSet 和 list的contains方法情况下，纯数组怎么去重？
可以这么做：
使用一个标识数组来标识重复的元素位置，再新建一个新的数组来存储去重后的元素
User user1 = new User("zjamgs",16,12); //User 需要重写equal 和 hashcode 方法
User user2 = new User("lisi",13,18);
User user3 = new User("wangwi",15,13);
User user4 = new User("zhangsli",23,10);
User user5 = new User("zjamgs",16,12);
User user6 = new User("zjamgs1",16,12);
User user7 = new User("lisi2",13,18);
User user8 = new User("zjamgs",16,12);
User user9 = new User("wangwi",15,13);
User user10 = new User("zjamgs5",16,12);
User[] users = {user1,user2,user3,user4,user5,user6,user7,user8,user9,user10};
int[] userindex = new int[users.length];//标记相同的元素的，0表示未重复，1表示已重复
int repeatCount = 0;
for(int i=0;i<users.length;i++){
if(userindex[i]==1)continue; //非常重要
for(int j=users.length-1;j>i;j--){
if(users[i].equals(users[j])){
userindex[j] = 1;
repeatCount++;
}
}
}
System.out.println(Arrays.toString(userindex);
User[] newUsers = new User[users.length - repeatCount];
for (int i = 0;i < userindex.length;i++){
if (userindex[i] == 0){
for (int j = 0;j < newUsers.length;j++){
if(newUsers[j] == null){
newUsers[j] = users[i];
break;
}
}
}
}
System.out.println(Arrays.toString(newUsers));
User类：
class User implements Comparable<User>{
private String name;
private int age;
private int index;
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof User)) return false;
User user = (User) o;
return getAge() == user.getAge() &&
getIndex() == user.getIndex() &&
Objects.equals(getName(), user.getName());
}
@Override
public int hashCode() {
return Objects.hash(getName(), getAge(), getIndex());
}
@Override
public String toString() {
return "User [name=" + name + ", age=" + age + ", index=" + index + "]";
}
public User(String name, int age, int index) {
super();
this.name = name;
this.age = age;
this.index = index;
}
public int getIndex() {
return index;
}
public void setIndex(int index) {
this.index = index;
}
public User() {
super();
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public int compareTo(User o) {
if(this.getAge()<o.getAge()){
return -1;
}else if(this.getAge()>o.getAge()){
return 1;
}
return 0;
}
}
输出结果：
[0, 0, 0, 0, 1, 0, 0, 1, 1, 0]
[User [name=zjamgs, age=16, index=12],
User [name=lisi, age=13, index=18],
User [name=wangwi, age=15, index=13],
User [name=zhangsli, age=23, index=10],
User [name=zjamgs1, age=16, index=12],
User [name=lisi2, age=13, index=18],
User [name=zjamgs5, age=16, index=12]]
有三个元素是前面已经有的元素，是重复元素，余下的7个都不是重复元素，结果正确