java中删除数组中重复元素方法探讨

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.


代码如下:

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);
    }

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。

代码如下:

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"),
    };　

代码如下:

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：

代码如下:

@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。

代码如下:

@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;
    }

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：

代码如下:

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：

代码如下:

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;
    }
｝

测试结果：

代码如下:

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