java数组去重

1、背景

根据不同的业务逻辑，经常会遇到数组中存在多个重复元素的场合,总结了下数组的排序，留个记录。

2、实现方法

总结了四种方法，接下来进行展示

1、方法一

[java]  view plain  copy

 

1.               //数组去重方法一  
2. String[] array = {"a","b","c","c","d","e","e","e","a"};  
3. List<String> result = new ArrayList<>();  
4. boolean flag;  
5. for(int i=0;i<array.length;i++){  
6.     flag = false;  
7.     for(int j=0;j<result.size();j++){  
8.         if(array[i].equals(result.get(j))){  
9.             flag = true;  
10.             break;  
11.         }  
12.     }  
13.     if(!flag){  
14.         result.add(array[i]);  
15.     }  
16. }  
17. String[] arrayResult = (String[]) result.toArray(new String[result.size()]);  
18. System.out.println(Arrays.toString(arrayResult));  

先遍历原数组，然后遍历结束集，通过每个数组的元素和结果集中的元素进行比对，若相同则break。若不相同，则存入结果集。

两层循环进行遍历得出最终结果。

2、方法二

[java]  view plain  copy

 

1. //数组去重方法二  
2. String[] array = {"a","b","c","c","d","e","e","e","a"};  
3. List<String> list = new ArrayList<>();  
4. list.add(array[0]);  
5. for(int i=1;i<array.length;i++){  
6.     if(list.toString().indexOf(array[i]) == -1){  
7.             list.add(array[i]);  
8.     }  
9. }  
10. String[] arrayResult = (String[]) list.toArray(new String[list.size()]);  
11. System.out.println(Arrays.toString(arrayResult));  

通过使用indexOf方法进行判断结果集中是否存在了数组元素。

3、方法三

[java]  view plain  copy

 

1. //数组去重方法三  
2. String[] array = {"a","b","c","c","d","e","e","e","a"};  
3. List<String> list = new ArrayList<>();  
4. for(int i=0;i<array.length;i++){  
5.     for(int j=i+1;j<array.length;j++){  
6.         if(array[i] == array[j]){  
7.             j = ++i;  
8.         }  
9.     }  
10.     list.add(array[i]);  
11. }  
12. String[] arrayResult = (String[]) list.toArray(new String[list.size()]);  
13. System.out.println(Arrays.toString(arrayResult));  

嵌套循环，进行比较获取满足条件结果集。

4、方法四

[java]  view plain  copy

 

1.         //数组去重方法四  
2.         String[] array = {"a","b","c","c","d","e","e","e","a"};  
3.         Arrays.sort(array);  
4.         List<String> list = new ArrayList<>();  
5.         list.add(array[0]);  
6.         for(int i=1;i<array.length;i++){  
7.             if(!array[i].equals(list.get(list.size()-1))){  
8.                 list.add(array[i]);  
9.             }  
10.         }  
11. <pre name="code" class="java"><span style="white-space:pre">        </span>String[] arrayResult = (String[]) list.toArray(new String[list.size()]);  

System.out.println(Arrays.toString(arrayResult));

先使用java提供的数组排序方法进行排序，然后进行一层for循环，进行相邻数据的比较即可获得最终结果集。

5、方法五

[java]  view plain  copy

 

1. //数组去重方法五  
2.         String[] array = {"a","b","c","c","d","e","e","e","a"};  
3.         Set<String> set = new HashSet<>();  
4.         for(int i=0;i<array.length;i++){  
5.             set.add(array[i]);  
6.         }  
7.         String[] arrayResult = (String[]) set.toArray(new String[set.size()]);  
8.         System.out.println(Arrays.toString(arrayResult));  

感谢 漂泊一剑客  的提议，加入set方法进行添加，虽然是无序排列，但是也更方便的解决了去重的问题。

3、知识说明

1、ArrayList集合转数组

[java]  view plain  copy

 

1. String[] arrayResult = (String[]) list.toArray(new String[list.size()]);  

对应的java方法API

toArray

public Object[] toArray()

Returns an array containing all of the elements in this list in proper sequence (from first to last element).

The returned array will be "safe" in that no references to it are maintained by this list. (In other words, this method must allocate a new array). The caller is thus free to modify the returned array.

This method acts as bridge between array-based and collection-based APIs.

Specified by:

toArray in interface  Collection<E>

Specified by:

toArray in interface  List<E>

Overrides:

toArray in class  AbstractCollection<E>

Returns:

an array containing all of the elements in this list in proper sequence

See Also:

Arrays.asList(Object[])

---

toArray

public <T> T[] toArray(T[] a)

Returns an array containing all of the elements in this list in proper sequence (from first to last element); the runtime type of the returned array is that of the specified array. If the list fits in the specified array, it is returned therein. Otherwise, a new array is allocated with the runtime type of the specified array and the size of this list.

If the list fits in the specified array with room to spare (i.e., the array has more elements than the list), the element in the array immediately following the end of the collection is set to null. (This is useful in determining the length of the list only if the caller knows that the list does not contain any null elements.)

Specified by:

toArray in interface  Collection<E>

Specified by:

toArray in interface  List<E>

Overrides:

toArray in class  AbstractCollection<E>

Parameters:

a - the array into which the elements of the list are to be stored, if it is big enough; otherwise, a new array of the same runtime type is allocated for this purpose.

Returns:

an array containing the elements of the list

Throws:

ArrayStoreException - if the runtime type of the specified array is not a supertype of the runtime type of every element in this list

NullPointerException - if the specified array is null

2、数组直接打印到控制台

直接调用Arrays的toString方法进行转换再进行打印操作。

实例：

[java]  view plain  copy

 

1. System.out.println(Arrays.toString(arrayResult));  

4、总结

仅仅是根据自己想法进行总结，肯定还有更多更优的方法能够去实现，希望大神指出教导。