java list去重操作實作方式

Java中的List是可以包含重複元素的(hash code 和equals)，那麼對List進行去重操作有兩種方式實現： 
方案一：可以透過HashSet來實現，程式碼如下： 

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; 
} 
}

必須實作hashCode和equals兩個方法，一會我們會看為啥必須實現 
具體的操作代碼如下： 

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())); 
}

調用代碼： 

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

利用HashSet進行去重操作，為啥必須覆蓋hashCode和equals兩個方法呢？
我們查看HashSet的add操作源碼如下： 

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

調用了HashMap進行操作的，我們看HashMap的put操作： 

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; 
}

需要注意的是： 

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

也就是說hash 。 
複雜度：一邊遍歷即可，O(n) 
方案二：直接遍歷一遍List進行透過contains和add作業實現 
程式碼如下： 

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。 
複雜度： 
一邊遍歷，同時呼叫了contains方法，我們查看原始碼如下： 

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; 
}

可以看到又對新的list做了一次遍歷操作。也就是1+2+....+n這樣複雜度為O(n*n) 
結論： 
方案一效率高，即採用HashSet的方式進行去重操作

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