基於JDK1.8的LinkedList原始碼學習筆記
阿新 • • 發佈:2019-01-15
LinkedList作為一種常用的List,是除了ArrayList之外最有用的List。其同樣實現了List介面,但是除此之外它同樣實現了Deque介面,而Deque是一個雙端佇列介面,其繼承自Queue,所以LinkedList同樣可以用來模擬佇列,棧以及雙端佇列。
一.基本用法
因為LinkedList是基於連結串列實現的,所以註定其插入和刪除操作速度要快於ArrayList,但是由於其是連結串列結構,所以其隨機訪問查詢檢索速度慢於基於陣列的ArrayList。
這裡先主要說一下LinkedList的基本用法,以及模擬佇列,模擬棧,模擬雙端佇列的常用方法。
1.LinkedList,List用法
List<String> myList=new LinkedList<String>(); (1)//增加元素 String s="myString" myList.add(s);//這裡等同於在連結串列尾端增加元素addLast(e) myList.add(1,s);//在指定位置插入元素 (2)//獲取指定位置的元素 String getString=myList.get(10)//獲取連結串列第11處元素,從頭計算 (3)//刪除元素 myList.remove(2)//刪除連結串列第3個元素 (4)//clear清空連結串列myList.clear() (5)isEmpty(),//判斷list是否為空
2.LinkedList模擬佇列
Queue<String> myQueue=new LinkedList<String>(); (1)//新增元素到到隊尾 myQueue.offer(myString); myQueue.add(myString); (2)檢索但不刪除隊首元素 String head=myQueue.peek();//若為空,返回null String head=myQueue.element();//若佇列為空,丟擲NoSuchElementException(3)取出並且刪除隊首元素 String head=myQueue.poll(); //若為空,返回null String head=myQueue.remove();//若佇列為空,丟擲NoSuchElementException //綜上,LinkedList通過在連結串列尾插入元素,連結串列首取出元素,模擬了先進先出FIFO的佇列,但是 //這裡的佇列是單向的
3.LinkedList模擬棧Stack操作
Deque<String> stack=new LinkedList<String>(); //(1)進棧操作 stack.push(myString); //(2)出棧操作,刪除並且取出 stack.pop(); //(3)若是檢索不刪除則還用peek stack.peek(); //LinkedList通過在隊首插入元素,隊首取出元素,模擬stack的先進後出操作
4.LinkedList模擬雙端佇列Deque操作
Deque<String> deque=new LinkedList<String>(); //(1)隊首新增元素 deque.offerFirst(myString); deque.addFirst(myString); //(2)隊尾新增元素 deque.offerLast(myString); deque.addLast(myString); //(3)檢索但不刪除隊首元素 String first=deque.peekFirst(); first=deque.getFirst(); //(4)檢索但不刪除隊尾元素 String last=deque.peekLast(); last=deque.getLast(); //(5)取出並刪除隊首元素 deque.pollFirst(); deque.removeFirst(); //(6)取出並刪除隊尾元素 deque.pollLast(); deque.removeLast(); //這樣LinkedList通過操作連結串列隊首隊尾就實現了雙端佇列
5.LinkedList迭代遍歷
//(1)for each 迴圈 List<String> list=new ArrayList<String>(); for(String s:list){ //// } //(2)iterator迭代器 Iterator<String> it=list.iterator(); while(it.hasNext()){ it.next(); } //(3)同時List還提供了ListIterator介面,擁有反向正向迭代 ListIterator<String> lit=list.listIterator(); while(lit.hasNext()){ it.next(); }//正向迭代 while(it.hasPrevious()){ it.previous(); }//反向迭代 //值得注意的是,以前可能忽視了,listIterator迭代器同時提供了增刪改的功能 //add(),在指定位置插入一個元素,當前迭代的前面插入 //set(E,e),修改當前迭代為指定元素 //remove();刪除上一次迭代
二.JDK原始碼分析
這裡的JDK是基於JDK1.8的原始碼。
1.定義,LinkedList類定義
public class LinkedList<E> extends AbstractSequentialList<E> implements List<E>, Deque<E>, Cloneable, java.io.Serializable // 繼承了AbstractSequentialList抽象類,提供了實現List介面的基本實現 //Deque介面, A linear collection that supports element insertion and removal at both ends. //The name <i>deque</i> is short for "double ended queue" and is usually pronounced "deck" public interface Deque<E> extends Queue<E> //所以這裡就可以知道為什麼LinkedList可以模擬佇列,雙端佇列,以及Stack棧了
2.重要屬性
transient int size = 0;//記錄List大小 //接下來分別是兩個Node引用,分別指向連結串列頭和連結串列尾 transient Node<E> first; transient Node<E> last; //接下就是連結串列中節點的定義,可以看到JDK1.8把節點都統一為Node了 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; } } //及其簡單的定義,雙向連結串列,向前連結,向後向後連結,元素
3.構造器
//(1)無參構造器 public LinkedList() { } //(2)帶有集合的構造器 public LinkedList(Collection<? extends E> c) { this(); addAll(c); } //呼叫addAll將現有集合內所有元素放到LinkedList中 public boolean addAll(Collection<? extends E> c) { return addAll(size, c); } //將整個集合c中的元素加入連結串列中 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; //插入到結尾 if (index == size) { succ = null; pred = last; } else {//插入到中間 //這裡succ則為原來在index位置的節點 succ = node(index); pred = succ.prev; } for (Object o : a) { @SuppressWarnings("unchecked") E e = (E) o; //建立新的Node節點,其中newNode的前向節點為pred,後向節點沒有定義 Node<E> newNode = new Node<>(pred, e, null); //pred==null,則此節點為首節點 if (pred == null) first = newNode; else //當節點不是首節點時,定義前向節點的後向節點為當前節點 pred.next = newNode; pred = newNode; } if (succ == null) { last = pred; } else { //將原來的連結串列加入 pred.next = succ; succ.prev = pred; } size += numNew; modCount++; return true; }
4.常用方法原始碼分析
(1). add(E e)
//預設add方法,將節點放入連結串列尾部,同offer方法 public boolean add(E e) { linkLast(e); return true; } //將節點放入連結串列尾部 void linkLast(E e) { final Node<E> l = last; final Node<E> newNode = new Node<>(l, e, null); last = newNode; //同樣要判斷當前節點是不是頭節點 if (l == null) first = newNode; else l.next = newNode; size++; modCount++; }
//將元素連結放到指定位置 public void add(int index, E element) { //該方法主要是檢視index是否合法,在範圍內,否則丟擲異常 checkPositionIndex(index); //當index是末尾時,直接連結到結尾 if (index == size) linkLast(element); else //否則找到index位置的原來節點,插入到其前面 linkBefore(element, node(index)); } //取出index位置的node節點 Node<E> node(int index) { // assert isElementIndex(index); //這裡有一處非常值得注意 //size>>1表示的是向右移位1,該方法其實相當於除以2,去得一半的值 //當index<size/2時,表明index在前半部分,則正序找 //否則在後半部分,則倒序查詢,節省了時間 if (index < (size >> 1)) { Node<E> x = first; for (int i = 0; i < index; i++) x = x.next; return x; } else { Node<E> x = last; for (int i = size - 1; i > index; i--) x = x.prev; return x; } } //linkBefore 方法 //這個方法是將節點插入到succ節點的前面, //由於是在指定位置插入節點,所以要將原來的節點連結到新節點後面 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 //這裡一定要注意,雙向連結串列,一定要將pred節點的next節點定義為當前節點 pred.next = newNode; size++; modCount++; }
(2).addLast(),addFirst()方法
addLast()等同於add()方法,addFirst是在連結串列頭插入節點
//將新節點放入到連結串列尾部 public void addLast(E e) { linkLast(e); } //在連結串列頭插入節點 public void addFirst(E e) { linkFirst(e); } //將新節點設定為首節點 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++; }
(3). getFirst(),getLast()獲取頭節點和尾節點
/** * Returns the first element in this list. * * @return the first element in this list * @throws NoSuchElementException if this list is empty為空會丟擲異常 */ public E getFirst() { final Node<E> f = first; if (f == null) throw new NoSuchElementException(); return f.item; } /** * 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; }
(4). removeFirst(),removeLast()方法
/** * Removes and returns the first element from this list. * * @return the first element from this list * @throws NoSuchElementException if this list is empty */
public E removeFirst() { final Node<E> f = first; if (f == null) throw new NoSuchElementException(); return unlinkFirst(f); } //unlinkFirst()即解開並返回頭節點 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;//及時清除 f.next = null; // help GC first = next; if (next == null) last = null;//此時連結串列為空 else next.prev = null; size--; modCount++; return element; }
/** * Removes and returns the last element from this list. * * @return the last element from this list * @throws NoSuchElementException if this list is empty */ 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; }
(5). contains(Object o)
檢視連結串列中是否存有某個元素
public boolean contains(Object o) { return indexOf(o) != -1; } //indexOf()這個方法返回物件O在連結串列中的位置 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 { //同樣呼叫的也是equals方法判斷兩個值是否相等 for (Node<E> x = first; x != null; x = x.next) { if (o.equals(x.item)) return index; index++; } } return -1;//沒有找到時返回-1 }
(6). get(int index)
獲取指定index位置的元素
/** * Returns the element at the specified position in this list. * * @param index index of the element to return * @return the element at the specified position in this list * @throws IndexOutOfBoundsException {@inheritDoc} */ public E get(int index) { checkElementIndex(index); return node(index).item; }
(7).set(int index,E element)
set修改指定位置的元素
//主要還是定位獲取節點之後再修改 public E set(int index, E element) { checkElementIndex(index); Node<E> x = node(index); E oldVal = x.item; x.item = element; return oldVal; }
(8).搜尋元素所在位置indexOf(Object o),lastIndexOf(Object o)
分為正向indexOf(),即第1次插入時匹配的元素位置和反向lastIndexOf(),即最後一次插入匹配的位置
//indexOf()這個方法返回物件O在連結串列中的位置 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 { //同樣呼叫的也是equals方法判斷兩個值是否相等 for (Node<E> x = first; x != null; x = x.next) { if (o.equals(x.item)) return index; index++; } } return -1; } //反向查詢 //有index的時候,必然會有lastIndexOf 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) { //這裡值得注意的是,index先--,因為你是從size位置開始的,所以要先-- index--; if (o.equals(x.item)) return index; } } return -1; }
5.模擬Queue操作原始碼分析
再次強調一次這裡queue先進先出,在隊尾入隊,隊首出隊
(1).首先是檢索隊首,但不出隊的操作,peek(),element()
/** * Retrieves, but does not remove, the head (first element) of this list. * * @return the head of this list, or {@code null} if this list is empty * @since 1.5 */最常用操作,peek(),若為空會,返回null 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 *///若為空會丟擲異常 public E element() { return getFirst(); } //再回頭看一眼getFirst(), public E getFirst() { final Node<E> f = first; if (f == null) throw new NoSuchElementException();//丟擲異常 return f.item; }
(2).出隊操作,取出隊首元素,poll(),remove()
/** * 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 */ public E remove() { return removeFirst(); } public E removeFirst() { final Node<E> f = first; if (f == null) throw new NoSuchElementException(); return unlinkFirst(f); }
(3).隊尾插入元素offer()
/** * 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); }
6. 模擬雙端佇列Deque操作原始碼分析
雙端佇列,其實就是整條連結串列頭尾都操作,有了前面的基礎,這裡應該非常簡單了
(1).在隊首,隊尾插入元素,offerFirst(),offerLast()
其實就是分別呼叫addFirst(E e)和addLast(E e)方法
/** * 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; }
(2).檢索隊首,隊尾元素,但不出隊peekFirst(),peekLast()
/** * 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; }