1. 重写的原因

如有个User对象如下：

public class User {
    private String name;
    private Integer age;
}

如果不重写equals方法和hashcode方法，则：

public static void main(String[] args) {
     User user1 = new User("userA", 30);
     User user2 = new User("userB", 30);
     System.out.println(user1.equals(user2)); // false
     System.out.println(user1.hashCode() == user2.hashCode()); // false
     System.out.println(user1.hashCode());
     System.out.println(user2.hashCode());
 }

2. 重写方法

equals:

@Override
public boolean equals(Object obj) {
    if (null == obj) {
        return false;
    }
    if (obj == this) {
        return true;
    }
    if (!(obj instanceof User)) {
        return false;
    }
    User anoUser = (User) obj;
    if (Objects.equals(this.name, anoUser.name) && Objects.equals(this.age, anoUser.age)) {
        return true;
    }
    return false;
}

hashcode:

@Override
public int hashCode() {
    return Objects.hash(this.name, this.age);
}

// 或
@Override
public int hashCode() {
    int result = 1;
    if (this.name == null && this.age == null) {
        return result;
    }
    result = 31 * result + (this.name == null ? 0 : this.name.hashCode());
    result = 31 * result + (this.age == null ? 0 : this.age.hashCode());
    return result;
}

如果hashcode返回一个固定值，则HashMap每次都要找同一个位置，导致链表很长，效率很低。