解码LinkedList实现

Question

只是为了理解实现,操作等,我从JDK复制了LinkedList类,并以非泛型(String only)的方式进行了修改.你可以在这里找到实际的代码.与问题相关,这里是
addLast()

public void addLast(String e){
    System.out.println("Invoked:addLast()");
    linkLast(e);
}

节点内部类

private static class Node {
     String item;
     Node next;
     Node prev;
     Node(Node prev, String element, Node next) {
         this.item = element;
         this.next = next;
         this.prev = prev;
    }
    @Override   public String toString() {
        return "Node [item=" + item + ", next=" + next + ", prev=" + prev+ "]";
    }
}

linkLast()

void linkLast(String e) {
    System.out.println("Invoked:linkLast(" + e + ")");
    final Node l = last;
    final Node newNode = new Node(l, e, null);
    System.out.println("\tCreatd new node as:" + newNode);
    last = newNode;
    System.out.println("\tSetting this node as last");
    if (l == null){
        // Applicable for 1st node. Last & First point to newNode
        first = newNode;
        System.out.println("\tLast was null, setting this node as first.");
    }          
    else{
        // Set newNode as next of previous Last.
        System.out.println("\tLast was not null, seting this node as next node of previous Last.");
        l.next = newNode;
    }
    size++;
    //System.out.println("Size of list:" + size);
    modCount++;
    System.out.println("\tMod Count:" + modCount + "\n");
}

我得到一个stackoverflow错误,当我执行这个类时,

public static void main(String args[]){
    LinkedListDemo obj = new LinkedListDemo();
    obj.add("B");
    obj.addFirst("A");
    obj.addLast("C");
}

产量

Invoked:add(B)
Invoked:linkLast(B)
    Creatd new node as:Node [item=B, next=null, prev=null]
    Setting this node as last   Last was null, setting this node as first.
    Mod Count:1
Invoked:addFirst()
Invoked:linkFirst(A)
    Creatd new node asNode [item=A, next=Node [item=B, next=null, prev=null], prev=null]
    Seting this node as first
    First was not null, seting this node as next node of previous First.
    Mod Count:2
Invoked:addLast()
Invoked:linkLast(C)
Exception in thread "main" java.lang.StackOverflowError
    at java.lang.String.getChars(String.java:826)
    at java.lang.AbstractStringBuilder.append(AbstractStringBuilder.java:416)at java.lang.StringBuilder.append(StringBuilder.java:132)
    at java.lang.StringBuilder.<init>(StringBuilder.java:110)
    at collections.LinkedListDemo$Node.toString(LinkedListDemo.java:570)
    at java.lang.String.valueOf(String.java:2847)
    at java.lang.StringBuilder.append(StringBuilder.java:128)
    at collections.LinkedListDemo$Node.toString(LinkedListDemo.java:570)

然后最后3行不断重复.我无法弄清楚造成这种情况的原因.

更新 - 基于收到的回复
所以,如果理解是正确的,

Invoked:add(A)
Invoked:linkLast(A)
    Creatd new node as:Node [item=A, next=null, prev=null]
    Setting this node as last
    Last was null, setting this node as first.
    Mod Count:1
Invoked:add(B)
Invoked:linkLast(B)
    Creatd new node as:Node [item=B, next=null, prev=Node [item=A, next=null, prev=null]]
    Setting this node as last
    Last was not null, seting this node as next node of previous Last.
    Mod Count:2

在,此阶段"节点B"仅使用单向链接(prev)创建,其中调用toString().只有在其他侧链接(下一个)被链接之后.并且这个双链接是下次插入时无限递归的原因.

也许这就是为什么toString()方法既没有在它的两个超类AbstractSequentialList,AbstractList中实现,也没有在AbstractCollection中实现的原因.该实现提取Iterator,迭代打印colletcion.

那么,为什么它只在第3次插入时发生.两次插入不会导致此问题.

Answer 1

问题是你的toString方法,它在尝试打印prev和next节点时递归调用它自己.

如果next打印的当前的Node不为空,next.toString()将被调用,而当prev那Node被打印出来,toString()原来Node被再次调用,所以没有结束递归.

可能的方法:

@Override
public String toString() 
{
    return "Node [item=" + item + ", next=" + (next==null?"null":next.item) + ", prev=" + (prev==null?"null":prev.item) + "]";
}

关于意见:

添加第一个节点后,列表如下所示:

prev             next
null <- Node1 -> null

first并且last都引用Node1

toString中没有无限递归的风险.

添加第二个节点后,列表如下所示:

null <- Node1 -> <- Node2 -> null

first是指Node1并且last指的是Node2

如您所见,Node1和Node2相互引用(Node1是Node2 prev,Node2是Node1 next).因此,尝试toString()在第二次插入后使用原始方法打印其中一个节点将导致无限递归.

但是,由于在l.next = newNode;执行行之前打印新节点,因此在第二次插入时不会获得无限递归,仅在第三次插入中.如果移动System.out.println("\tCreatd new node as:" + newNode);到结尾linkLast,第二次插入将导致无限递归.