问题：比如我有一个数组（元素个数为0哈），希望添加进去元素不能重复。
拿到这样一个问题，我可能会快速的写下代码，这里数组用ArrayList.
private static void testListSet(){
List arrays = new ArrayList(){
@Override
public boolean add(String e) {
for(String str:this){
if(str.equals(e)){
System.out.println("add failed !!! duplicate element");
return false;
}else{
System.out.println("add successed !!!");
}
}
return super.add(e);
}
};
arrays.add("a");arrays.add("b");arrays.add("c");arrays.add("b");
for(String e:arrays)
System.out.print(e);
}
这里我什么都不关，只关心在数组添加元素的时候做下判断（当然添加数组元素只用add方法），是否已存在相同元素，如果数组中不存在这个元素，就添加到这个数组中，反之亦然。这样写可能简单，但是面临庞大数组时就显得笨拙：有100000元素的数组天家一个元素，难道要调用100000次equal吗？这里是个基础。
问题：加入已经有一些元素的数组了，怎么删除这个数组里重复的元素呢？
大家知道java中集合总的可以分为两大类：List与Set。List类的集合里元素要求有序但可以重复，而Set类的集合里元素要求无序但不能重复。那么这里就可以考虑利用Set这个特性把重复元素删除不就达到目的了，毕竟用系统里已有的算法要优于自己现写的算法吧。
public static void removeDuplicate(List list){
HashSet set = new HashSet(list);
list.clear();
list.addAll(set);
} private static People[] ObjData = new People[]{
new People(0, "a"),new People(1, "b"),new People(0, "a"),new People(2, "a"),new People(3, "c"),
};
public class People{
private int id;
private String name;
public People(int id,String name){
this.id = id;
this.name = name;
}
@Override
public String toString() {
return ("id = "+id+" , name "+name);
}
}
上面的代码，用了一个自定义的People类，当我添加相同的对象时候（指的是含有相同的数据内容），调用removeDuplicate方法发现这样并不能解决实际问题，仍然存在相同的对象。那么HashSet里是怎么判断像个对象是否相同的呢？打开HashSet源码可以发现：每次往里面添加数据的时候，就必须要调用add方法：
@Override
public boolean add(E object) {
return backingMap.put(object, this) == null;
}
这里的backingMap也就是HashSet维护的数据，它用了一个很巧妙的方法，把每次添加的Object当作HashMap里面的KEY，本身HashSet对象当作VALUE。这样就利用了Hashmap里的KEY唯一性，自然而然的HashSet的数据不会重复。但是真正的是否有重复数据，就得看HashMap里的怎么判断两个KEY是否相同。
@Override public V put(K key, V value) {
if (key == null) {
return putValueForNullKey(value);
}
int hash = secondaryHash(key.hashCode());
HashMapEntry[] tab = table;
int index = hash & (tab.length - 1);
for (HashMapEntry e = tab[index]; e != null; e = e.next) {
if (e.hash == hash && key.equals(e.key)) {
preModify(e);
V oldValue = e.value;
e.value = value;
return oldValue;
}
}
// No entry for (non-null) key is present; create one
modCount++;
if (size++ > threshold) {
tab = doubleCapacity();
index = hash & (tab.length - 1);
}
addNewEntry(key, value, hash, index);
return null;
}
总的来说，这里实现的思路是：遍历hashmap里的元素，如果元素的hashcode相等（事实上还要对hashcode做一次处理），然后去判断KEY的eqaul方法。如果这两个条件满足，那么就是不同元素。那这里如果数组里的元素类型是自定义的话，要利用Set的机制，那就得自己实现equal与hashmap（这里hashmap算法就不详细介绍了，我也就理解一点）方法了：
public class People{
private int id; //
private String name;
public People(int id,String name){
this.id = id;
this.name = name;
}
@Override
public String toString() {
return ("id = "+id+" , name "+name);
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public boolean equals(Object obj) {
if(!(obj instanceof People))
return false;
People o = (People)obj;
if(id == o.getId()&&name.equals(o.getName()))
return true;
else
return false;
}
@Override
public int hashCode() {
// TODO Auto-generated method stub
return id;
//return super.hashCode();
}
}
这里在调用removeDuplicate(list)方法就不会出现两个相同的people了。
好吧，这里就测试它们的性能吧：
public class RemoveDeplicate {
public static void main(String[] args) {
// TODO Auto-generated method stub
//testListSet();
//removeDuplicateWithOrder(Arrays.asList(data));
//ArrayList list = new ArrayList(Arrays.asList(ObjData));
//removeDuplicate(list);
People[] data = createObjectArray(10000);
ArrayList list = new ArrayList(Arrays.asList(data));
long startTime1 = System.currentTimeMillis();
System.out.println("set start time --> "+startTime1);
removeDuplicate(list);
long endTime1 = System.currentTimeMillis();
System.out.println("set end time --> "+endTime1);
System.out.println("set total time --> "+(endTime1-startTime1));
System.out.println("count : " + People.count);
People.count = 0;
long startTime = System.currentTimeMillis();
System.out.println("Efficient start time --> "+startTime);
EfficientRemoveDup(data);
long endTime = System.currentTimeMillis();
System.out.println("Efficient end time --> "+endTime);
System.out.println("Efficient total time --> "+(endTime-startTime));
System.out.println("count : " + People.count);
}
public static void removeDuplicate(List list)
{
HashSet set = new HashSet(list);
list.clear();
list.addAll(set);
}
public static void removeDuplicateWithOrder(List arlList)
{
Set set = new HashSet();
List newList = new ArrayList();
for (Iterator iter = arlList.iterator(); iter.hasNext();) {
String element = iter.next();
if (set.add( element))
newList.add( element);
}
arlList.clear();
arlList.addAll(newList);
}
@SuppressWarnings("serial")
private static void testListSet(){
List arrays = new ArrayList(){
@Override
public boolean add(String e) {
for(String str:this){
if(str.equals(e)){
System.out.println("add failed !!! duplicate element");
return false;
}else{
System.out.println("add successed !!!");
}
}
return super.add(e);
}
};
arrays.add("a");arrays.add("b");arrays.add("c");arrays.add("b");
for(String e:arrays)
System.out.print(e);
}
private static void EfficientRemoveDup(People[] peoples){
//Object[] originalArray; // again, pretend this contains our original data
int count =0;
// new temporary array to hold non-duplicate data
People[] newArray = new People[peoples.length];
// current index in the new array (also the number of non-dup elements)
int currentIndex = 0;
// loop through the original array...
for (int i = 0; i < peoples.length; ++i) {
// contains => true iff newArray contains originalArray[i]
boolean contains = false;
// search through newArray to see if it contains an element equal
// to the element in originalArray[i]
for(int j = 0; j 26
count : 3653
Efficient start time --> 1326443326729
efficient medthod inner count : 28463252
Efficient end time --> 1326443327107
Efficient total time --> 378
count : 28463252