关于Java Collections常见的10个问题

下面是在StackOverflow中常问及和讨论的Java集合最流行问题。在你看这些问题前，最好先看一下 类层次结构图.

1. When to use LinkedList over ArrayList?

ArrayList is essentially an array. Its elements can be accessed directly by index. But if the array is full, a new larger array is needed to allocate and moving all elements to the new array will take O(n) time. Also adding or removing an element needs to move existing elements in an array. This might be the most disadvantage to use ArrayList.

LinkedList is a double linked list. Therefore, to access an element in the middle, it has to search from the beginning of the list. On the other hand, adding and removing an element in LinkedList is quicklier, because it only changes the list locally.

In summary, the worst case of time complexity comparison is as follows:

                   | Arraylist | LinkedList   ------------------------------------------   get(index)        |    O(1)   |   O(n)   add(E)            |    O(n)   |   O(1)   add(E, index)     |    O(n)   |   O(n)   remove(index)     |    O(n)   |   O(n)   Iterator.remove() |    O(n)   |   O(1)   Iterator.add(E)   |    O(n)   |   O(1)

Despite the running time, memory usage should be considered too especially for large lists. In LinkedList, every node needs at least two extra pointers to link the previous and next nodes; while in ArrayList, only an array of elements is needed.

More comparisons between list.

2. 当迭代集合时高效删除元素

The only correct way to modify a collection while iterating is using Iterator.remove(). For example,

Iterator<Integer> itr = list.iterator();  while(itr.hasNext()) {     // do something     itr.remove();  }

One most frequent incorrect code is

for(Integer i: list) {    list.remove(i);  }

You will get a ConcurrentModificationException by running the above code. This is because an iterator has been generated (in for statement) to traverse over the list, but at the same time the list is changed by Iterator.remove(). In Java, "it is not generally permissible for one thread to modify a collection while another thread is iterating over it."

3. 如何将List 转成 int[]?

The easiest way might be using ArrayUtils in Apache Commons Lang library.

int[] array = ArrayUtils.toPrimitive(list.toArray(new Integer[0]));

In JDK, there is no short-cut. Note that you can not use List.toArray(), because that will convert List to Integer[]. The correct way is following,

int[] array = new int[list.size()];  for(int i=0; i < list.size(); i++) {    array[i] = list.get(i);  }

4. 如何将 int[] 数组转成 List?

The easiest way might still be using ArrayUtils in Apache Commons Lang library, like below.

List list = Arrays.asList(ArrayUtils.toObject(array));

In JDK, there is no short-cut either.

int[] array = {1,2,3,4,5};  List<Integer> list = new ArrayList<Integer>();  for(int i: array) {    list.add(i);  }

5. 什么是过滤一个集合的最好方式?

Again, you can use third-party package, like Guava or Apache Commons Lang to fullfil this function. Both provide a filter() method (in Collections2 of Guava and in CollectionUtils of Apache). The filter() method will return elements that match a given Predicate.

In JDK, things become harder. A good news is that in Java 8, Predicate will be added. But for now you have to use Iterator to traverse the whole collection.

Iterator<Integer> itr = list.iterator();  while(itr.hasNext()) {     int i = itr.next();     if (i > 5) { // filter all ints bigger than 5        itr.remove();     }  }

Of course you can mimic the way of what Guava and Apache did, by introducing a new interface Predicate. That might also be what most advanced developers will do.

public interface Predicate<T> {     boolean test(T o);  }     public static <T> void filter(Collection<T> collection, Predicate<T> predicate) {      if ((collection != null) && (predicate != null)) {         Iterator<T> itr = collection.iterator();            while(itr.hasNext()) {              T obj = itr.next();              if (!predicate.test(obj)) {                 itr.remove();              }          }      }  }

Then we can use the following code to filter a collection:

filter(list, new Predicate<Integer>() {      public boolean test(Integer i) {          return i <= 5;       }  });

6. 最简单的方式将一个List转成Set?

There are two ways to do so, depending on how you want equal defined. The first piece of code puts a list into a HashSet. Duplication is then identified mostly by hashCode(). In most cases, it will work. But if you need to specify the way of comparison, it is better to use the second piece of code where you can define your own comparator.

Set<Integer> set = new HashSet<Integer>(list);

Set<Integer> set = new TreeSet<Integer>(aComparator);  set.addAll(list);

7. 如何移除ArrayList中重复的元素？

This question is quite related to the above one.
If you don't care the ordering of the elements in the ArrayList, a clever way is to put the list into a set to remove duplication, and then to move it back to the list. Here is the code

ArrayList** list = ... // initial a list with duplicate elements  Set<Integer> set = new HashSet<Integer>(list);  list.clear();  list.addAll(set);

If you DO care about the ordering, order can be preserved by putting a list into a LinkedHashSet which is in the standard JDK.

8. 排序集合

There are a couple of ways to maintain a sorted collection in Java. All of them provide a collection in natural ordering or by the specified comparator. By natural ordering, you also need to implement the Comparable interface in the elements.

1. Collections.sort() can sort a List. As specified in the javadoc, this sort is stable, and guarantee n log(n) performance.
2. PriorityQueue provides an ordered queue. The difference between PriorityQueue and Collections.sort() is that, PriorityQueue maintain an order queue at all time, but you can only get the head element from the queue. You can not randomly access its element like PriorityQueue.get(4).
3. If there is no duplication in the collection, TreeSet is another choice. Same as PriorityQueue, it maintains the ordered set at all time. You can get the lowest and highest elements from the TreeSet. But you still cannot randomly access its element.

In a short, Collections.sort() provides a one-time ordered list. PriorityQueue and TreeSet maintain ordered collections at all time, in the cost of no indexed access of elements.

9. Collections.emptyList() vs new instance

The same question applies to emptyMap() and emptySet().

Both methods return an empty list, but Collections.emptyList() returns a list that is immutable. This mean you cannot add new elements to the "empty" list. At the background, each call of Collections.emptyList() actually won't create a new instance of an empty list. Instead, it will reuse the existing empty instance. If you are familiar Singleton in the design pattern, you should know what I mean. So this will give you better performance if called frequently.

10 Collections.copy集合复制

There are two ways to copy a source list to a destination list. One way is to use ArrayList constructor

ArrayList<Integer> dstList = new ArrayList<Integer>(srcList);

The other is to use Collections.copy() (as below). Note the first line, we allocate a list at least as long as the source list, because in the javadoc of Collections, it says The destination list must be at least as long as the source list.

ArrayList<Integer> dstList = new ArrayList<Integer>(srcList.size());  Collections.copy(dstList, srcList);

Both methods are shallow copy. So what is the difference between these two methods?

• First, Collections.copy() won't reallocate the capacity of dstList even if dstList does not have enough space to contain all srcList elements. Instead, it will throw an IndexOutOfBoundsException. One may question if there is any benefit of it. One reason is that it guarantees the method runs in linear time. Also it makes suitable when you would like to reuse arrays rather than allocate new memory in the constructor of ArrayList.

• Collections.copy() can only accept List as both source and destination, while ArrayList accepts Collection as the parameter, therefore more general.