List in Java can contain repeated elements (hash code and equals), so there are two ways to deduplicate List:
Option 1: It can be implemented through HashSet. The code is as follows:
Copy the code code 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;
}
}
Two methods, hashCode and equals, must be implemented. We will see in a moment why the specific operation codes must be implemented as follows:
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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:
Copy the code code as follows:
public static void main(String[] args) {
removeListDuplicateObject();
}
When using HashSet to perform deduplication operations, why do we need to override the hashCode and equals methods?
Let’s check the source code of HashSet’s add operation as follows:
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public boolean add(E e) {
return map.put(e, PRESENT)==null;
}
HashMap is called for operation. Let's look at the put operation of HashMap:
Copy the code code as follows:
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;
}
Things to note are:
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if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
...
}
In other words, the hash codes are equal and equals(==).
Complexity: Just traverse on one side, O(n)
Option 2: Directly traverse the List and implement the contains and add operations. The code is as follows:
Copy the code code 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()));
}
Others are the same as above.
Complexity:
While traversing, the contains method is called at the same time. We view the source code as follows:
Copy the code 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 another traversal operation has been performed on the new list. That is, 1+2+....+n, the complexity is O(n*n)
in conclusion:
Solution 1 is highly efficient, that is, using HashSet to perform deduplication operations.
