java list去重操作实现方式
Java中的List是可以包含重复元素的(hash code 和equals),那么对List进行去重操作有两种方式实现:
方案一:可以通过HashSet来实现,代码如下:
代码如下:
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;
}
}
必须实现hashCode和equals两个方法,一会我们会看为啥必须实现
具体的操作代码如下:
代码如下:
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()));
}
public static void main(String[] args) {
removeListDuplicateObject();
}
利用HashSet进行去重操作,为啥必须覆盖hashCode和equals两个方法呢?
我们查看HashSet的add操作源码如下:
代码如下:
public boolean add(E e) {
return map.put(e, PRESENT)==null;
}
调用了HashMap进行操作的,我们看HashMap的put操作:
代码如下:
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;
}
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
......
}
也就是说hash code相等且equals(==)。
复杂度:一边遍历即可,O(n)
方案二:直接遍历一遍List进行通过contains和add操作实现
代码如下:
代码如下:
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()));
}
其他等同上面。
复杂度:
一边遍历,同时调用了contains方法,我们查看源码如下:
代码如下:
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;
}
可以看到又对新的list做了一次遍历操作。也就是1+2+....+n这样复杂度为O(n*n)
结论:
方案一效率高,即采用HashSet的方式进行去重操作