java中刪除數組中反復元素辦法商量。本站提示廣大學習愛好者：（java中刪除數組中反復元素辦法商量）文章只能為提供參考，不一定能成為您想要的結果。以下是java中刪除數組中反復元素辦法商量正文
成績：好比我有一個數組（元素個數為0哈），願望添加出來元素不克不及反復。
拿到如許一個成績，我能夠會疾速的寫下代碼，這裡數組用ArrayList.
private static void testListSet(){
List<String> arrays = new ArrayList<String>(){
@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);
}
這裡我甚麼都不關，只關懷在數組添加元素的時刻做下斷定（固然添加數組元素只用add辦法），能否已存在雷同元素，假如數組中不存在這個元素，就添加到這個數組中，反之亦然。如許寫能夠簡略，然則面對宏大數組時就顯得愚笨：有100000元素的數組天家一個元素，豈非要挪用100000次equal嗎？這裡是個基本。
成績：參加曾經有一些元素的數組了，怎樣刪除這個數組裡反復的元素呢？
年夜家曉得java中聚集總的可以分為兩年夜類：List與Set。List類的聚集裡元素請求有序但可以反復，而Set類的聚集裡元素請求無序但不克不及反復。那末這裡便可以斟酌應用Set這個特征把反復元素刪除不就到達目標了，究竟用體系裡已有的算法要優於本身現寫的算法吧。
public static void removeDuplicate(List<People> list){
HashSet<People> set = new HashSet<People>(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);
}
}
下面的代碼，用了一個自界說的People類，當我添加雷同的對象時刻（指的是含有雷同的數據內容），挪用removeDuplicate辦法發明如許其實不能處理現實成績，依然存在雷同的對象。那末HashSet裡是怎樣斷定像個對象能否雷同的呢？翻開HashSet源碼可以發明：每次往外面添加數據的時刻，就必需要挪用add辦法：
@Override
public boolean add(E object) {
return backingMap.put(object, this) == null;
}
這裡的backingMap也就是HashSet保護的數據，它用了一個很奇妙的辦法，把每次添加的Object看成HashMap外面的KEY，自己HashSet對象看成VALUE。如許就應用了Hashmap裡的KEY獨一性，天然而然的HashSet的數據不會反復。然則真實的能否有反復數據，就得看HashMap裡的怎樣斷定兩個KEY能否雷同。
@Override public V put(K key, V value) {
if (key == null) {
return putValueForNullKey(value);
}
int hash = secondaryHash(key.hashCode());
HashMapEntry<K, V>[] tab = table;
int index = hash & (tab.length - 1);
for (HashMapEntry<K, V> 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;
}
總的來講，這裡完成的思緒是：遍歷hashmap裡的元素，假如元素的hashcode相等（現實上還要對hashcode做一次處置），然後去斷定KEY的eqaul辦法。假如這兩個前提知足，那末就是分歧元素。那這裡假如數組裡的元素類型是自界說的話，要應用Set的機制，那就得本身完成equal與hashmap（這裡hashmap算法就不具體引見了，我也就懂得一點）辦法了：
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();
}
}
這裡在挪用removeDuplicate(list)辦法就不會湧現兩個雷同的people了。
好吧，這裡就測試它們的機能吧：
public class RemoveDeplicate {
public static void main(String[] args) {
// TODO Auto-generated method stub
//testListSet();
//removeDuplicateWithOrder(Arrays.asList(data));
//ArrayList<People> list = new ArrayList<People>(Arrays.asList(ObjData));
//removeDuplicate(list);
People[] data = createObjectArray(10000);
ArrayList<People> list = new ArrayList<People>(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<People> list)
{
HashSet<People> set = new HashSet<People>(list);
list.clear();
list.addAll(set);
}
public static void removeDuplicateWithOrder(List<String> arlList)
{
Set<String> set = new HashSet<String>();
List<String> newList = new ArrayList<String>();
for (Iterator<String> 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<String> arrays = new ArrayList<String>(){
@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 <= currentIndex; ++j) {
// if the same element is found, don't add it to the new array
count++;
if(peoples[i].equals(newArray[j])) {
contains = true;
break;
}
}
// if we didn't find a duplicate, add the new element to the new array
if(!contains) {
// note: you may want to use a copy constructor, or a .clone()
// here if the situation warrants more than a shallow copy
newArray[currentIndex] = peoples[i];
++currentIndex;
}
}
System.out.println("efficient medthod inner count : "+ count);
}
private static People[] createObjectArray(int length){
int num = length;
People[] data = new People[num];
Random random = new Random();
for(int i = 0;i<num;i++){
int id = random.nextInt(10000);
System.out.print(id + " ");
data[i]=new People(id, "i am a man");
}
return data;
}
｝
測試成果：
set end time --> 1326443326724
set total time --> 26
count : 3653
Efficient start time --> 1326443326729
efficient medthod inner count : 28463252
Efficient end time --> 1326443327107
Efficient total time --> 378
count : 28463252
