Java list deduplication implementation

The List in Java can contain repeated elements (hash code and equals), so there are two ways to implement the rework operation on the List:
Solution 1: it can be realized by HashSet, the code is as follows:

 
class Student { 
private String id; 
private String name; 
public Student(String id, String name) { 
super(); 
this.id = id; 
this.name = name; 
} 
@Override 
public String toString() { 
return "Student [id=" + id + ", name=" + name + "]"; 
} 
@Override 
public int hashCode() { 
final int prime = 31; 
int result = 1; 
result = prime * result + ((id == null) ? 0 : id.hashCode()); 
result = prime * result + ((name == null) ? 0 : name.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; 
} 
Student other = (Student) obj; 
if (id == null) { 
if (other.id != null) { 
return false; 
} 
} else if (!id.equals(other.id)) { 
return false; 
} 
if (name == null) { 
if (other.name != null) { 
return false; 
} 
} else if (!name.equals(other.name)) { 
return false; 
} 
return true; 
} 
} 

HashCode and equals must be implemented, and we'll see why in a moment
The specific operation code is as follows:

 
private static void removeListDuplicateObject() { 
List<Student> list = new ArrayList<Student>(); 
for (int i = 0; i < 10; i++) { 
Student student = new Student("id", "name"); 
list.add(student); 
} 
System.out.println(Arrays.toString(list.toArray())); 
Set<Student> set = new HashSet<Student>(); 
set.addAll(list); 
System.out.println(Arrays.toString(set.toArray())); 
list.removeAll(list); 
set.removeAll(set); 
System.out.println(Arrays.toString(list.toArray())); 
System.out.println(Arrays.toString(set.toArray())); 
} 

Calling code:

 
public static void main(String[] args) { 
removeListDuplicateObject(); 
} 

Why do you have to override hashCode and equals for a HashSet deduplication operation?
We see the add operation of HashSet source code as follows:

 
public boolean add(E e) { 
return map.put(e, PRESENT)==null; 
} 

Call the operation of HashMap, let's look at the put operation of HashMap:

 
public V put(K key, V value) { 
if (key == null) 
return putForNullKey(value); 
int hash = hash(key.hashCode()); 
int i = indexFor(hash, table.length); 
for (Entry<K,V> e = table[i]; e != null; e = e.next) { 
Object k; 
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) { 
V oldValue = e.value; 
e.value = value; 
e.recordAccess(this); 
return oldValue; 
} 
} 
modCount++; 
addEntry(hash, key, value, i); 
return null; 
} 

Note:

 
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) { 
...... 
} 

That is, hash code is equal and equals(==).
Complexity: traversal on one side, O(n)
Option 2: iterate over the List directly through the contains and add operations
The code is as follows:

 
private static void removeListDuplicateObjectByList() { 
List<Student> list = new ArrayList<Student>(); 
for (int i = 0; i < 10; i++) { 
Student student = new Student("id", "name"); 
list.add(student); 
} 
System.out.println(Arrays.toString(list.toArray())); 
List<Student> listUniq = new ArrayList<Student>(); 
for (Student student : list) { 
if (!listUniq.contains(student)) { 
listUniq.add(student); 
} 
} 
System.out.println(Arrays.toString(listUniq.toArray())); 
list.removeAll(list); 
listUniq.removeAll(listUniq); 
System.out.println(Arrays.toString(list.toArray())); 
System.out.println(Arrays.toString(listUniq.toArray())); 
} 

The others are equivalent to the top.
Complexity:
While traversing, we also called the contains method, we look at the source code as follows:

 
public boolean contains(Object o) { 
return indexOf(o) >= 0; 
} 
public int indexOf(Object o) { 
if (o == null) { 
for (int i = 0; i < size; i++) 
if (elementData[i]==null) 
return i; 
} else { 
for (int i = 0; i < size; i++) 
if (o.equals(elementData[i])) 
return i; 
} 
return -1; 
} 

You can see that the new list is traversed again. That is 1 + 2 +... Plus n is order n times n.
Conclusion:
Scheme 1 is efficient, that is, the method of HashSet to carry out the deduplication operation