Collection
ArrayList源码剖析
添加元素分析
ArrayList数组源码剖析
三个构造器
代码语言：javascript复制public ArrayList(int initialCapacity) {
if (initialCapacity > 0) {
this.elementData = new Object[initialCapacity];
} else if (initialCapacity == 0) {
this.elementData = EMPTY_ELEMENTDATA;
} else {
throw new IllegalArgumentException("Illegal Capacity: "
initialCapacity);
}
}
/**
* Constructs an empty list with an initial capacity of ten.
*/
public ArrayList() {
this.elementData = DEFAULTCAPACITY_EMPTY_ELEMENTDATA;
}
/**
* Constructs a list containing the elements of the specified
* collection, in the order they are returned by the collection's
* iterator.
*
* @param c the collection whose elements are to be placed into this list
* @throws NullPointerException if the specified collection is null
*/
//将另一个数组直接全部赋值给新的数组，然后初始化
public ArrayList(Collection<? extends E> c) {
elementData = c.toArray();
if ((size = elementData.length) != 0) {
// c.toArray might (incorrectly) not return Object[] (see 6260652)
if (elementData.getClass() != Object[].class)
elementData = Arrays.copyOf(elementData, size, Object[].class);
} else {
// replace with empty array.
this.elementData = EMPTY_ELEMENTDATA;
}
}对于图中的拷贝如下图
add方法
他不仅可以实现添加，也可以实现插入操作
代码语言：javascript复制public boolean add(E e) {
ensureCapacityInternal(size 1); // Increments modCount!!
elementData[size ] = e;
return true;
}
/**
* Inserts the specified element at the specified position in this
* list. Shifts the element currently at that position (if any) and
* any subsequent elements to the right (adds one to their indices).
*
* @param index index at which the specified element is to be inserted
* @param element element to be inserted
* @throws IndexOutOfBoundsException {@inheritDoc}
*/
public void add(int index, E element) {
rangeCheckForAdd(index);
ensureCapacityInternal(size 1); // Increments modCount!!
System.arraycopy(elementData, index, elementData, index 1,
size - index);
elementData[index] = element;
size ;
}以及对应的addAll方法,也可以是实现插入操作
代码语言：javascript复制public boolean addAll(Collection<? extends E> c) {
Object[] a = c.toArray();
int numNew = a.length;
ensureCapacityInternal(size numNew); // Increments modCount
System.arraycopy(a, 0, elementData, size, numNew);
size = numNew;
return numNew != 0;
}
/**
* Inserts all of the elements in the specified collection into this
* list, starting at the specified position. Shifts the element
* currently at that position (if any) and any subsequent elements to
* the right (increases their indices). The new elements will appear
* in the list in the order that they are returned by the
* specified collection's iterator.
*
* @param index index at which to insert the first element from the
* specified collection
* @param c collection containing elements to be added to this list
* @return <tt>true</tt> if this list changed as a result of the call
* @throws IndexOutOfBoundsException {@inheritDoc}
* @throws NullPointerException if the specified collection is null
*/
public boolean addAll(int index, Collection<? extends E> c) {
rangeCheckForAdd(index);
Object[] a = c.toArray();
int numNew = a.length;
ensureCapacityInternal(size numNew); // Increments modCount
int numMoved = size - index;
if (numMoved > 0)
System.arraycopy(elementData, index, elementData, index numNew,
numMoved);
System.arraycopy(a, 0, elementData, index, numNew);
size = numNew;
return numNew != 0;
}Set方法
对于set方法，就像源码中提及的那样，如果判断索引位置数组下标没有越界，那么就直接赋值即可
代码语言：javascript复制public E set(int index, E element) {
rangeCheck(index);//下标越界检查
E oldValue = elementData(index);
elementData[index] = element;//赋值到指定位置，复制的仅仅是引用
return oldValue;
}get方法
获取数据索引位置数据，检查是否越界。如果没有，然后返回即可
代码语言：javascript复制public E get(int index) {
rangeCheck(index);
return (E) elementData[index];//注意类型转换
}remove方法
传入要删除位置的索引，然后将索引位置后面的元素全部向前挪一位即可（赋值拷贝）
代码语言：javascript复制public E remove(int index) {
rangeCheck(index);
modCount ;
E oldValue = elementData(index);
int numMoved = size - index - 1;
if (numMoved > 0)
System.arraycopy(elementData, index 1, elementData, index, numMoved);
elementData[--size] = null; //清除该位置的引用，让GC起作用
return oldValue;
}trimToSize()方法
作用 : 将底层数组的容量调整为当前列表保存的实际元素的大小的功能。 ,以减少内存的开销
/**
* Trims the capacity of this <tt>ArrayList</tt> instance to be the
* list's current size. An application can use this operation to minimize
* the storage of an <tt>ArrayList</tt> instance.
*/
public void trimToSize() {
modCount ;
if (size < elementData.length) {
elementData = (size == 0)
? EMPTY_ELEMENTDATA
: Arrays.copyOf(elementData, size);
}
}indexOf()—获取元素第一次出现的位置
代码语言：javascript复制/**
* Returns the index of the first occurrence of the specified element
* in this list, or -1 if this list does not contain the element.
* More formally, returns the lowest index <tt>i</tt> such that
* <tt>(o==null ? get(i)==null : o.equals(get(i)))</tt>,
* or -1 if there is no such index.
*/
public int indexOf(Object o) {
//因为List可以存储null对象，所以源码地方在这里将其考虑进去，分开进行索引
if (o == null) {
for (int i = 0; i < size; i )
if (elementData[i]==null)
return i;
} else {
for (int i = 0; i < size; i )
if (o.equals(elementData[i]))
return i;
}
return -1;
}lastIndexOf() — 获取元素最后一次出现的位置
从最后开始索引，然后出现的第一个就是最后一个。源码牛
代码语言：javascript复制/**
* Returns the index of the last occurrence of the specified element
* in this list, or -1 if this list does not contain the element.
* More formally, returns the highest index <tt>i</tt> such that
* <tt>(o==null ? get(i)==null : o.equals(get(i)))</tt>,
* or -1 if there is no such index.
*/
public int lastIndexOf(Object o) {
if (o == null) {
for (int i = size-1; i >= 0; i--)
if (elementData[i]==null)
return i;
} else {
for (int i = size-1; i >= 0; i--)
if (o.equals(elementData[i]))
return i;
}
return -1;
}LinkedList源码剖析
LinkedList<E> res = new LinkedList<>();
LinkedList同时实现了List接口和Deque接口，所以说我们既可以将其视为一个容器，又可以将其视为一个队列乃至栈
LinkedList底层通过双向链表实现
LinkedList通过first和last引用分别指向链表的第一个和最后一个元素。注意这里没有所谓的哑元，当链表为空的时候first和last都指向null。
LinkedList它的内部定义了一个私有的Node链表
代码语言：javascript复制private static class Node<E> {
E item;
Node<E> next;
Node<E> prev;
Node(Node<E> prev, E element, Node<E> next) {
this.item = element;
this.next = next;
this.prev = prev;
}
}
三个构造器
transient是Java语言的关键字，用来表示一个成员变量不是该对象序列化的一部分
public LinkedList() {
}
/**
* Constructs a list containing the elements of the specified
* collection, in the order they are returned by the collection's
* iterator.
*
* @param c the collection whose elements are to be placed into this list
* @throws NullPointerException if the specified collection is null
*/
public LinkedList(Collection<? extends E> c) {
this();
addAll(c);
}
/**
* Links e as first element.
*/
private void linkFirst(E e) {
final Node<E> f = first;
final Node<E> newNode = new Node<>(null, e, f);
first = newNode;
if (f == null)
last = newNode;
else
f.prev = newNode;
size ;
modCount ;
}
/**
* Links e as last element.
*/
void linkLast(E e) {
final Node<E> l = last;
final Node<E> newNode = new Node<>(l, e, null);
last = newNode;
if (l == null)
first = newNode;
else
l.next = newNode;
size ;
modCount ;
}getFirst() 获取第一个元素
只要对LinkedList一设置值，那么last 和first就会被赋值
代码语言：javascript复制/**
* Returns the first element in this list.
*
* @return the first element in this list
* @throws NoSuchElementException if this list is empty
*/
public E getFirst() {
final Node<E> f = first;
if (f == null)
throw new NoSuchElementException();
return f.item;
}getLast() 获取最后一个元素:
代码语言：javascript复制/**
* Returns the last element in this list.
*
* @return the last element in this list
* @throws NoSuchElementException if this list is empty
*/
public E getLast() {
final Node<E> l = last;
if (l == null)
throw new NoSuchElementException();
return l.item;
}删除元素（如下）
removeFirst(), removeLast(), remove(e), remove(index)
从源码中我们可以得出,如果需要删除某个节点，那么就将这个节点传入，或者传入该节点的索引
代码语言：javascript复制 public boolean remove(Object o) {
if (o == null) {
for (Node<E> x = first; x != null; x = x.next) {
if (x.item == null) {
unlink(x);
return true;
}
}
} else {
for (Node<E> x = first; x != null; x = x.next) {
if (o.equals(x.item)) {
unlink(x);
return true;
}
}
}
return false;
}
E unlink(Node<E> x) {
// assert x != null;
final E element = x.item;
final Node<E> next = x.next;
final Node<E> prev = x.prev;
if (prev == null) {// 第一个元素
first = next;
} else {
prev.next = next;
x.prev = null;
}
if (next == null) {// 最后一个元素
last = prev;
} else {
next.prev = prev;
x.next = null;
}
x.item = null; // GC
size--;
modCount ;
return element;
}
对于removeFirst 和 removeLast
代码语言：javascript复制public E removeFirst() {
final Node<E> f = first;
if (f == null)
throw new NoSuchElementException();
return unlinkFirst(f);
}
/**
* Unlinks non-null first node f.
*/
private E unlinkFirst(Node<E> f) {
// assert f == first && f != null;
final E element = f.item;
final Node<E> next = f.next;
f.item = null;
f.next = null; // help GC
first = next;
if (next == null)
last = null;
else
next.prev = null;
size--;
modCount ;
return element;
}
public E removeLast() {
final Node<E> l = last;
if (l == null)
throw new NoSuchElementException();
return unlinkLast(l);
}
/**
* Unlinks non-null last node l.
*/
private E unlinkLast(Node<E> l) {
// assert l == last && l != null;
final E element = l.item;
final Node<E> prev = l.prev;
l.item = null;
l.prev = null; // help GC
last = prev;
if (prev == null)
first = null;
else
prev.next = null;
size--;
modCount ;
return element;
}
add方法
对于Add方法，官网给出了两类添加的方法，一种是直接添加到链表的末尾。还有一种是插入链表。
代码语言：javascript复制public boolean add(E e) {
linkLast(e);
return true;
}
/**
* Links e as last element.
*/
void linkLast(E e) {
final Node<E> l = last;
final Node<E> newNode = new Node<>(l, e, null);
last = newNode;
if (l == null)
first = newNode;
else
l.next = newNode;
size ;
modCount ;
}
public void add(int index, E element) {
//先检查索引位置是否越界
checkPositionIndex(index);
if (index == size)
linkLast(element);
else
linkBefore(element, node(index));
}
void linkBefore(E e, Node<E> succ) {
// assert succ != null;
final Node<E> pred = succ.prev;
final Node<E> newNode = new Node<>(pred, e, succ);
succ.prev = newNode;
if (pred == null)
first = newNode;
else
pred.next = newNode;
size ;
modCount ;
}
addAll()
同样，addAll也是有两种构造器，第一个是将另一个集合全部添加到链表的末尾，另一个是将另一个集合添加到指定的位置
代码语言：javascript复制public boolean addAll(Collection<? extends E> c) {
return addAll(size, c);
}
/**
* Inserts all of the elements in the specified collection into this
* list, starting at the specified position. Shifts the element
* currently at that position (if any) and any subsequent elements to
* the right (increases their indices). The new elements will appear
* in the list in the order that they are returned by the
* specified collection's iterator.
*
* @param index index at which to insert the first element
* from the specified collection
* @param c collection containing elements to be added to this list
* @return {@code true} if this list changed as a result of the call
* @throws IndexOutOfBoundsException {@inheritDoc}
* @throws NullPointerException if the specified collection is null
*/
public boolean addAll(int index, Collection<? extends E> c) {
checkPositionIndex(index);
Object[] a = c.toArray();
int numNew = a.length;
if (numNew == 0)
return false;
Node<E> pred, succ;
if (index == size) {
succ = null;
pred = last;
} else {
succ = node(index);
pred = succ.prev;
}
for (Object o : a) {
@SuppressWarnings("unchecked") E e = (E) o;
Node<E> newNode = new Node<>(pred, e, null);
if (pred == null)
first = newNode;
else
pred.next = newNode;
pred = newNode;
}
if (succ == null) {
last = pred;
} else {
pred.next = succ;
succ.prev = pred;
}
size = numNew;
modCount ;
return true;
}Set方法
对于set方法，就像源码中提及的那样，如果判断索引位置数组下标没有越界，那么就直接赋值即可
代码语言：javascript复制/**
* Replaces the element at the specified position in this list with the
* specified element.
*
* @param index index of the element to replace
* @param element element to be stored at the specified position
* @return the element previously at the specified position
* @throws IndexOutOfBoundsException {@inheritDoc}
*/
public E set(int index, E element) {
checkElementIndex(index);
Node<E> x = node(index);
E oldVal = x.item;
x.item = element;
return oldVal;
}get方法
获取数据索引位置数据，检查是否越界。如果没有，然后返回即可
代码语言：javascript复制public E get(int index) {
checkElementIndex(index);
return node(index).item;
}
Clear()
代码语言：javascript复制/**
* Removes all of the elements from this list.
* The list will be empty after this call returns.
*/
public void clear() {
// Clearing all of the links between nodes is "unnecessary", but:
// - helps a generational GC if the discarded nodes inhabit
// more than one generation
// - is sure to free memory even if there is a reachable Iterator
for (Node<E> x = first; x != null; ) {
Node<E> next = x.next;
x.item = null;
x.next = null;
x.prev = null;
x = next;
}
first = last = null;
size = 0;
modCount ;
}查看元素（indexOf / lastIndexOf）
和ArrayList集合一样，indexOf作用就是为了返回元素首次出现的位置
而lastIndexOf就是返回元素最后出现的位置
代码语言：javascript复制public int indexOf(Object o) {
int index = 0;
if (o == null) {
for (Node<E> x = first; x != null; x = x.next) {
if (x.item == null)
return index;
index ;
}
} else {
for (Node<E> x = first; x != null; x = x.next) {
if (o.equals(x.item))
return index;
index ;
}
}
return -1;
}
/**
* Returns the index of the last occurrence of the specified element
* in this list, or -1 if this list does not contain the element.
* More formally, returns the highest index {@code i} such that
* <tt>(o==null ? get(i)==null : o.equals(get(i)))</tt>,
* or -1 if there is no such index.
*
* @param o element to search for
* @return the index of the last occurrence of the specified element in
* this list, or -1 if this list does not contain the element
*/
public int lastIndexOf(Object o) {
int index = size;
if (o == null) {
for (Node<E> x = last; x != null; x = x.prev) {
index--;
if (x.item == null)
return index;
}
} else {
for (Node<E> x = last; x != null; x = x.prev) {
index--;
if (o.equals(x.item))
return index;
}
}
return -1;
}Queue<Integer> queue = new LinkedList<>();
使用接口 对象名 = new 类名; 方式实例化的对象只能调用接口中有的方法，而不能调用类中特有的方法。
而使用类名 对象名 = new 类名;方式创建出来的对象可以调用所有的方法
对于LinkedList() 它同样可以实现队列
方法有
- peek() —– 返回队列的头部
- element() —–返回队列的头部【 此方法与 peek 的不同之处仅在于，如果此队列为空，它会抛出异常。】
- poll() ——出列
- offer() —– 入列【使用容量受限队列时，此方法通常更可取， add后者只能通过引发异常来插入元素。】
- add() ——入列
- remove() —– 删除队列的头部
public E peek() {
final Node<E> f = first;
return (f == null) ? null : f.item;
}
/**
* Retrieves, but does not remove, the head (first element) of this list.
*
* @return the head of this list
* @throws NoSuchElementException if this list is empty
* @since 1.5
检索但不删除此队列的头部。 此方法与 peek 的不同之处仅在于，如果此队列为空，它会抛出异常。
*/
public E element() {
return getFirst();
}
/**
* Retrieves and removes the head (first element) of this list.
*
* @return the head of this list, or {@code null} if this list is empty
* @since 1.5
*/
public E poll() {
final Node<E> f = first;
return (f == null) ? null : unlinkFirst(f);
}
/**
* Retrieves and removes the head (first element) of this list.
*
* @return the head of this list
* @throws NoSuchElementException if this list is empty
* @since 1.5
检索并删除此队列的头部，如果此队列为空，则返回 null 。
返回值:
此队列的头部，或者 null 如果此队列为空
*/
public E remove() {
return removeFirst();
}
/**
* Adds the specified element as the tail (last element) of this list.
*
* @param e the element to add
* @return {@code true} (as specified by {@link Queue#offer})
* @since 1.5
*/
public boolean offer(E e) {
return add(e);
}
Deque双端队列
它在queue的基础上又增加了特殊情况，也就是返回值
此接口定义访问双端元素的方法。提供了插入、删除和检查元素的方法。这些方法中的每一个都以两种形式存在：一种在操作失败时引发异常，另一种返回特殊值（ null 或 false，具体取决于操作）。后一种形式的插入操作专门设计用于容量受限 Deque 的实现;在大多数实现中，插入操作不会失败。
详见源码
代码语言：javascript复制/**
* Inserts the specified element at the front of this list.
*
* @param e the element to insert
* @return {@code true} (as specified by {@link Deque#offerFirst})
* @since 1.6
*/
public boolean offerFirst(E e) {
addFirst(e);
return true;
}
/**
* Inserts the specified element at the end of this list.
*
* @param e the element to insert
* @return {@code true} (as specified by {@link Deque#offerLast})
* @since 1.6
*/
public boolean offerLast(E e) {
addLast(e);
return true;
}
/**
* Retrieves, but does not remove, the first element of this list,
* or returns {@code null} if this list is empty.
*
* @return the first element of this list, or {@code null}
* if this list is empty
* @since 1.6
*/
public E peekFirst() {
final Node<E> f = first;
return (f == null) ? null : f.item;
}
/**
* Retrieves, but does not remove, the last element of this list,
* or returns {@code null} if this list is empty.
*
* @return the last element of this list, or {@code null}
* if this list is empty
* @since 1.6
*/
public E peekLast() {
final Node<E> l = last;
return (l == null) ? null : l.item;
}
/**
* Retrieves and removes the first element of this list,
* or returns {@code null} if this list is empty.
*
* @return the first element of this list, or {@code null} if
* this list is empty
* @since 1.6
*/
public E pollFirst() {
final Node<E> f = first;
return (f == null) ? null : unlinkFirst(f);
}
/**
* Retrieves and removes the last element of this list,
* or returns {@code null} if this list is empty.
*
* @return the last element of this list, or {@code null} if
* this list is empty
* @since 1.6
*/
public E pollLast() {
final Node<E> l = last;
return (l == null) ? null : unlinkLast(l);
}
/**
* Pushes an element onto the stack represented by this list. In other
* words, inserts the element at the front of this list.
*
* <p>This method is equivalent to {@link #addFirst}.
*
* @param e the element to push
* @since 1.6
*/
public void push(E e) {
addFirst(e);
}
/**
* Pops an element from the stack represented by this list. In other
* words, removes and returns the first element of this list.
*
* <p>This method is equivalent to {@link #removeFirst()}.
*
* @return the element at the front of this list (which is the top
* of the stack represented by this list)
* @throws NoSuchElementException if this list is empty
* @since 1.6
*/
public E pop() {
return removeFirst();
}
/**
* Removes the first occurrence of the specified element in this
* list (when traversing the list from head to tail). If the list
* does not contain the element, it is unchanged.
*
* @param o element to be removed from this list, if present
* @return {@code true} if the list contained the specified element
* @since 1.6
*/
public boolean removeFirstOccurrence(Object o) {
return remove(o);
}
/**
* Removes the last occurrence of the specified element in this
* list (when traversing the list from head to tail). If the list
* does not contain the element, it is unchanged.
*
* @param o element to be removed from this list, if present
* @return {@code true} if the list contained the specified element
* @since 1.6
*/
public boolean removeLastOccurrence(Object o) {
if (o == null) {
for (Node<E> x = last; x != null; x = x.prev) {
if (x.item == null) {
unlink(x);
return true;
}
}
} else {
for (Node<E> x = last; x != null; x = x.prev) {
if (o.equals(x.item)) {
unlink(x);
return true;
}
}
}
return false;
