Java
Java-集合框架

源码解析-LinkedList

# 概述

LinkedList继承图如下 linkedlist1

可以看到LinkedList同时实现了List接口和Deque接口，也就是说它既可以看作一个顺序容器，又可以看作一个队列（Queue），同时又可以看作一个栈（Stack）。这样看来，LinkedList简直就是个全能冠军。当你需要使用栈或者队列时，可以考虑使用LinkedList，一方面是因为Java官方已经声明不建议使用Stack类，更遗憾的是，Java里根本没有一个叫做Queue的类（它是个接口名字）。关于栈或队列，现在的首选是ArrayDeque，它有着比LinkedList（当作栈或队列使用时）有着更好的性能。

LinkedList的实现方式决定了所有跟下标相关的操作都是线性时间，而在首段或者末尾删除元素只需要常数时间。为追求效率LinkedList没有实现同步（synchronized），如果需要多个线程并发访问，可以先采用Collections.synchronizedList()方法对其进行包装。

LinkedList 一些主要特性：

LinkedList 集合底层实现的数据结构为双向链表
LinkedList 集合中元素允许为 null
LinkedList 允许存入重复的数据
LinkedList 中元素存放顺序为存入顺序。
LinkedList 是非线程安全的，如果想保证线程安全的前提下操作 LinkedList，可以使用 List list = Collections.synchronizedList(new LinkedList(...)); 来生成一个线程安全的 LinkedList

# 底层数据结构

链表是一种不同于数组的数据结构，双向链表是链表的一种子数据结构，它具有以下的特点：

每个节点上有三个字段：当前节点的数据字段（data）,指向上一个节点的字段（prev），和指向下一个节点的字段（next）。

LinkedList底层通过双向链表实现，本节将着重讲解插入和删除元素时双向链表的维护过程，也即是直接跟List接口相关的函数，而将Queue和Stack以及Deque相关的知识放在下一节讲。双向链表的每个节点用内部类Node表示。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。当链表为空的时候first和last都指向null。

//Node内部类
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;
    }
}

1
2
3
4
5
6
7
8
9
10
11

linkedlist2

# 构造函数

//元素数量
transient int size = 0;

/**
 * Pointer to first node.   
 * Invariant: (first == null && last == null) ||
 *            (first.prev == null && first.item != null)
  指向整体链表的第一个元素
 */
transient Node<E> first;

/**
 * Pointer to last node.
 * Invariant: (first == null && last == null) ||
 *            (last.next == null && last.item != null)
 //指向链表的第二个元素
 */
transient Node<E> last;

/**
 * Constructs an empty list.
  无参构造器
 */
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
   传入一个集合的构造器
 */
public LinkedList(Collection<? extends E> c) {
    this();
    addAll(c);
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39

# 核心方法源码解析

# add()

add方法有两种，实现略微不同，一种是add(E e)，另外一个方法是在指定位置添加元素 add(int index E e)，我们先看 add(E e)。

add(E e)是在当前链表的最后端直接插入一个元素，源码比较简单，具体步骤参考注释

/**
 * Appends the specified element to the end of this list.
 *
 * <p>This method is equivalent to {@link #addLast}.
 *
 * @param e element to be appended to this list
 * @return {@code true} (as specified by {@link Collection#add})
 */
public boolean add(E e) {
  	//插入尾部
    linkLast(e);
  	//直接返回true
    return true;
}
/**
     * Links e as last element.
     */
void linkLast(E e) {
  //获取last节点，需要一个单独变量 赋值一下
  final Node<E> l = last;
  //把新插入的元素新建节点，具体参考 Node这个类的定义 l:前一个元素  e:当前元素 null:下一个元素
  final Node<E> newNode = new Node<>(l, e, null);
  //把last指向 新的元素节点
  last = newNode;
  //l == null表示 链表为空的情况下，需要顺便给first单独赋值,last已经在上一步赋值了
  if (l == null)
    first = newNode;
  else
    l.next = newNode;//链表不是空的情况下，需要把l的next指向当前的last(即新建的newNode节点),last的pre在new Node()的时候已经指向了l节点了
  size++;//当前存储的size+1
  modCount++; //修改次数+1
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33

add(int index,E e) 是在指定位置插入一个元素，实现方式是如果 index == size 那就跟add(E e)逻辑保持一致，如果index是中间某个坐标，那就是修改对应位置的node的前后节点的引用来达到目的，具体可以看代码注释。代码中的node(int index)函数有一点小小的trick，因为链表双向的，可以从开始往后找，也可以从结尾往前找，具体朝那个方向找取决于条件index < (size >> 1)，也即是index是靠近前端还是后端。从这里也可以看出，linkedList通过index检索元素的效率没有arrayList高。

/**
 * 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) {
  	//check下避免index超出整体 size范围
    checkPositionIndex(index);
		//如果插入的index是最后，那跟add(E e)逻辑保持一致
    if (index == size)
        linkLast(element);
    else
      	//插入中间某个节点
        linkBefore(element, node(index));
}

    /**
     * Returns the (non-null) Node at the specified element index.
       这个方法目的是返回指定index的node
     */
    Node<E> node(int index) {
        // assert isElementIndex(index);
        //比较trick的地方，如果index < size/2 那就从前往后遍历查找，否则从后往前遍历，也就是说 靠前就从前遍历，靠后就从后遍历，提升效率
        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;
        }
    }

    /**
     * Inserts element e before non-null Node succ.
     */
    void linkBefore(E e, Node<E> succ) {
        // assert succ != null;
      	//获取到 前置节点 
        final Node<E> pred = succ.prev;
        //新建需要插入的节点，pre指向前置节点，后置节点指向 找到的succ节点
        final Node<E> newNode = new Node<>(pred, e, succ);
        //succ的前置节点修改为 新节点
        succ.prev = newNode;
      	//插入位置为0 
        if (pred == null)
            first = newNode;
        else
            pred.next = newNode;  //前置节点的 next指向新节点
        size++;
        modCount++;
    }

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59

linkedlist3

# remove()

remove操作也分两种 remove(Object o),remove(int index)。remove(Object o)实现的是移除 o这个元素，remove(int index)实现的是移除指定index上的元素。实现比较简单，具体直接参考注释即可，这里给大家建议，链表的代码看起来会比较绕，可以对着代码在草稿本上画图来看整个代码逻辑会比较清晰。

/**
 * Removes the element at the specified position in this list.  Shifts any
 * subsequent elements to the left (subtracts one from their indices).
 * Returns the element that was removed from the list.
 *
 * @param index the index of the element to be removed
 * @return the element previously at the specified position
 * @throws IndexOutOfBoundsException {@inheritDoc}
 */
public E remove(int index) {
  	//check下index是否越界
    checkElementIndex(index);
    return unlink(node(index));
}

    /**
     * Returns the (non-null) Node at the specified element index.
       这个方法目的是返回指定index的node，具体注释参考 add方法中的介绍
     */
    Node<E> node(int index) {
        // assert isElementIndex(index);

        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;
        }
    }


    /**
     * Unlinks non-null node x.
      移除指定node。假设链表是 a -> node ->b ，这个方法做的就是把 a的next指向b，然后 b的pre指向a，最后形成 a->b ，中间很多代码都是各种特殊情况处理
     */
    E unlink(Node<E> x) {
        // assert x != null;
        final E element = x.item;
      	//获取 后置节点 next
        final Node<E> next = x.next;
      	//获取 前置节点 prev
        final Node<E> prev = x.prev;
				
        
        if (prev == null) {
          	//移除的是头部节点，需要把first变量重新赋值
            first = next;
        } else {
            //前置节点 next 指向next节点
            prev.next = next;
          	//解除 需要移除的节点的pre的联系
            x.prev = null;
        }
				
        if (next == null) {
          	//移除的节点是尾部节点，需要堆last重新赋值
            last = prev;
        } else {
            //后置节点的prev 指向 前置节点
            next.prev = prev;
            //解除需要移除的节点 next指向关系
            x.next = null;
        }
        //把x的item 赋为null，然后gc回收
        x.item = null;
      	//size -1
        size--;
      	//修改次数+1
        modCount++;
      	//返回被移除的元素
        return element;
    }


    /**
     * Removes the first occurrence of the specified element from this list,
     * if it is present.  If this list does not contain the element, it is
     * unchanged.  More formally, removes the element with the lowest index
     * {@code i} such that
     * <tt>(o==null&nbsp;?&nbsp;get(i)==null&nbsp;:&nbsp;o.equals(get(i)))</tt>
     * (if such an element exists).  Returns {@code true} if this list
     * contained the specified element (or equivalently, if this list
     * changed as a result of the call).
     *
     * @param o element to be removed from this list, if present
     * @return {@code true} if this list contained the specified element
     */
    public boolean remove(Object o) {
      		// 需要移除node的值为null的节点
        if (o == null) {
            for (Node<E> x = first; x != null; x = x.next) {
                if (x.item == null) {
                    //找到具体节点以后 移除
                    unlink(x);
                    return true;
                }
            }
        } else {
          
            for (Node<E> x = first; x != null; x = x.next) {
              	//节点比较是否相等 使用的equals
                if (o.equals(x.item)) {
                  	//找到具体节点以后 移除
                    unlink(x);
                    return true;
                }
            }
        }
        return false;
    }

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116

linkedlist4

# get()

实现比较简单，参考注释

/**
 * 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);
    //获取指定的元素 然后返回对应的值
    return node(index).item;
}

1
2
3
4
5
6
7
8
9
10
11
12

# set()

set需要注意的是set方法是值替换

/**
 * 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);
  	//找到对应index的节点
    Node<E> x = node(index);
  	//只替换了值,没有生成新的节点
    E oldVal = x.item;
    x.item = element;
    return oldVal;
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18

# clear()

clear做的事就是遍历链表的每一个元素然后把每一个元素全部置空，包括 first和last

/**
 * 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++;
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

# Queue相关的方法

LinkedList实现的队列的属性本质上还是调用已经实现的链表的相关方法，源码都比较简单就不讲解了，可以尝试自己使用LinkedList来实现一个Queue或者双向队列Deque（Double-end Queue），有非常多的面试题也是类似的要求。


    /**
     * 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
     */
    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
     */
    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
     */
    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);
    }

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56

# Deque相关方法

  /**
     * 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;
    }

    /**
     * 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;
    }

# 参考文献

https://github.com/CarpenterLee/JCFInternals/blob/master/markdown/3-LinkedList.md

https://pdai.tech/md/java/collection/java-collection-LinkedList.html#queue-%E6%96%B9%E6%B3%95

https://juejin.cn/post/6844903586095169549

上次更新: 2024/06/05, 22:36:57

← 源码解析-ArrayList 源码解析-Stack & Queue→