Question

问：链表的每个节点都有一个随机指针（除了下一个指针），它可以随机指向另一个节点或为空。你会如何复制这样的链表？

答：这就是我所拥有的，我只想批准这是否是最佳方式。

由于没有指定空间限制，我将使用LinkedHashSet和LinkedHashMap（我可以想象人们已经在分歧中点头;）

第一次迭代：显而易见 - 从列表中读取每个节点以进行复制并在新列表上创建节点。然后，像这样读取随机节点：this.random.data并插入LinkedHashSet。

第二次迭代：遍历新列表并将每个节点的数据添加为第一列，将节点本身作为第二列添加到LinkedHashMap中（不必链接，但我只是去随着流程。）

第三次迭代：迭代LinkedHashSet（这就是为什么需要链接 - 可预测的排序）和新列表同时进行。对于第一个节点，读取LinkedHashSet的第一个条目，在LinkedHashMap中查找相应的对象，并将随机节点添加到新列表中的当前节点。

3次迭代似乎有点疯狂，但尝试将复杂性保持为O（N）。任何改进O（3N）空间要求和O（3N）运行时复杂性的解决方案都会很棒。谢谢！

编辑：进入LinkedHashSet时，可以删除LinkedHashMap中的条目，因此只需要O（2N）空间。

Answer 1

作为pointed out by MahlerFive，我认为你可以用O（2N）运行时复杂度和O（N）空间复杂度来做到这一点。

我们假设你有

public class Node {
    private Node next;
    private Node random;
    private String data;
    // getters and setters omitted for the sake of brevity
}

我会对Node s的链接列表进行深度复制：

private Node deepCopy(Node original) {
    // We use the following map to associate newly created instances 
    // of Node with the instances of Node in the original list
    Map<Node, Node> map = new HashMap<Node, Node>();
    // We scan the original list and for each Node x we create a new 
    // Node y whose data is a copy of x's data, then we store the 
    // couple (x,y) in map using x as a key. Note that during this 
    // scan we set y.next and y.random to null: we'll fix them in 
    // the next scan
    Node x = original;
    while (x != null) {
        Node y = new Node();
        y.setData(new String(x.getData()));
        y.setNext(null);
        y.setRandom(null);
        map.put(x, y);
        x = x.getNext();
    }
    // Now for each Node x in the original list we have a copy y 
    // stored in our map. We scan again the original list and 
    // we set the pointers buildings the new list
    x = original;
    while (x != null) {
            // we get the node y corresponding to x from the map
        Node y = map.get(x);
            // let x' = x.next; y' = map.get(x') is the new node 
            // corresponding to x'; so we can set y.next = y'
        y.setNext(map.get(x.getNext()));
            // let x'' = x.random; y'' = map.get(x'') is the new 
            // node corresponding to x''; so we can set y.random = y''
        y.setRandom(map.get(x.getRandom()));
        x = x.getNext();
    }
    // finally we return the head of the new list, that is the Node y
    // in the map corresponding to the Node original
    return map.get(original);
}

修改：我发现此问题与问here的问题重复：您找到了answer，其中显示了如何在O（3N）中解决此问题）没有额外空间的运行时复杂性：非常巧妙！但它使用C指针的技巧，我不知道如何在java中做同样的事情。

Answer 2

您可以使用2N步骤和包含N个元素的地图执行此操作。

按照“下一个”指针行走旧列表。对于您访问的每个节点，将节点添加到新列表，将新列表中的上一个节点连接到新节点，将旧节点随机指针存储在新新节点中，然后将旧节点指针的映射存储到新节点中。地图中的新节点指针。
走新列表，并为每个随机指针在地图中查找，以找到新列表中的关联节点以替换它。

Answer 3

我最近在采访中也被问过这个问题。这是我提出的建议。创建原始列表节点的映射，其中每个节点的addreess将是键，随机指针的偏移量将是值。现在使用原始地图中的random pointer = null创建一个新的链表。最后，迭代原始列表，借助map get原始指针的偏移量，并使用该偏移量链接新创建的地图中的随机指针。

面试官最后并不高兴。可能正在寻找更好的方法，或者他心中已经找到了答案，无法掌握解决问题的新方法。

Answer 4

走list并使用clone()？

Answer 5

在O（n）时间和恒定空间

public class CloneLinkedListWithRandomPointer {

public static void main(String[] args) throws Exception {
    SpecialLink link = new SpecialLink(1);
    SpecialLink two = new SpecialLink(2);
    SpecialLink three = new SpecialLink(3);
    SpecialLink four = new SpecialLink(4);
    SpecialLink five = new SpecialLink(5);

    link.next = two;
    two.next = three;
    three.next = four;
    four.next = five;

    link.random = four;
    two.random = five;
    three.random = null;
    four.random = five;
    five.random=link;

    SpecialLink copy = cloneSpecialLinkedList(link);

    System.out.println(link);
    System.out.println(copy);
}


public static SpecialLink cloneSpecialLinkedList(SpecialLink link) throws Exception{

    SpecialLink temp = link;
    while(temp != null){
        temp.next = (SpecialLink) temp.clone();
        temp = temp.next==null?temp.next:temp.next.next;
    }

    temp = link;
    while(temp != null){
        temp.next.random = temp.random!=null?temp.random.next:null;
        temp = temp.next==null?temp.next:temp.next.next;
    }


    SpecialLink copy = link.next;

    temp = link;
    SpecialLink copyTemp = copy;

    while(temp.next!= null && copyTemp.next != null){
        temp.next = temp.next.next;
        copyTemp.next = copyTemp.next.next;

        temp = temp.next;
        copyTemp = copyTemp.next;

    }

    return copy;
}

}

class SpecialLink implements Cloneable{

enum Type{
    ORIGINAL,COPY
}

int val;
SpecialLink next;
SpecialLink random;
Type type;

public void setValue(int value){
    this.val = value;
}

public SpecialLink addNode(int value){
    return next = new SpecialLink(value);
}

public SpecialLink(int value) {
    super();
    this.val = value;
    this.type = Type.ORIGINAL;
}

@Override
public String toString() {
    SpecialLink temp = this;
    StringBuilder builder = new StringBuilder();
    while(temp != null){
        builder.append(temp.val).append("--").append(temp.type.toString()).append("->").append(temp.random == null? null:temp.random.val).append("--").append(temp.random == null? null:temp.random.type);
        builder.append(", ");
        temp = temp.next;
    }
    return builder.toString();
}

@Override
public Object clone() throws CloneNotSupportedException {
    // TODO Auto-generated method stub
    SpecialLink clone = (SpecialLink) super.clone();
    clone.type = Type.COPY;
    return clone;
}
}

Answer 6

我为@MahlerFive的解决方案编写了代码，该代码无需映射。

以下是代码：

private static class Node {
    private String item;
    private Node next;
    private Node random;
}

public static Node cloneLinkedStructure(Node head) {
    // make holes after each original node
    for (Node p = head; p != null;) {
        Node pnext = p.next;
        Node hole = new Node();
        hole.item = ".";
        p.next = hole;
        hole.next = pnext;
        p = pnext;
    }

    Node fakeHead = new Node(); // fake new head
    Node q = fakeHead;
    Node p = head;
    while (p != null) {
        // build the new linked structure
        Node oldq = q;
        q = new Node();
        q.item = p.item;
        oldq.next = q;
        q.random = p.random.next; // link to a hole

        Node hole = p.next;
        hole.random = q; // use link RANDOM as a backward link to new node

        p = hole.next;
    }
    q.next = null;

    Node newHead = fakeHead.next; // throw fake head
    // build random links for the new linked structure
    for (q = newHead; q != null; q = q.next)
        q.random = q.random.random;

    // delete holes to restore original linked structure
    for (p = head; p != null; p = p.next)
        p.next = p.next.next;

    return newHead;
}

Answer 7

1）创建节点1的副本并将其插入节点1和节点1之间。原始链接列表中的节点2，创建2的副本并将其插入2和2之间。 3 ..以这种方式继续，在第N个节点后添加N的副本

2）现在以这种方式复制任意链接
original->next->arbitrary = original->arbitrary->next;  /*TRAVERSE TWO NODES*/
这是有效的，因为原始＆gt;接下来只是原始副本和原始 - ＆gt;任意 - ＆gt;接下来只是任意的副本。 3）现在以这种方式在单个循环中恢复原始和复制链接列表。
original->next = original->next->next;
copy->next = copy->next->next;
4）确保original-＆gt; next的最后一个元素为NULL。

时间复杂度：O（n）

辅助空间：O（1）

source

Answer 8

这是Java实现：

public static <T> RandomLinearNode<T> clone(RandomLinearNode<T> head) {
    if (head == null) {
        return head;
    }
    RandomLinearNode<T> itr = head, temp;

    // insert copy nodes after each original nodes
    while (itr != null) {
        temp = new RandomLinearNode<T>(itr.getElement());
        temp.next(itr.next());
        itr.next(temp);
        itr = temp.next();
    }
    // copy the random pointer
    itr = head;
    while (itr != null && itr.next() != null) {
        if (itr.random() != null) {
            itr.next().random(itr.random().next());
        }
        itr = itr.next().next();
    }
    // break the list into two
    RandomLinearNode<T> newHead = head.next();
    itr = head;
    while (itr != null && itr.next() != null) {
        temp = itr.next();
        itr.next(temp.next());          
        itr = temp.next();
    }
    return newHead;
}

这是单元测试

@Test
public void cloneLinkeListWithRandomPointerTest() {
    RandomLinearNode<Integer> one = new RandomLinearNode<Integer>(1, null, null);
    RandomLinearNode<Integer> two = new RandomLinearNode<Integer>(2, one, null);
    RandomLinearNode<Integer> three = new RandomLinearNode<Integer>(3, two, null);
    RandomLinearNode<Integer> four = new RandomLinearNode<Integer>(4, three, null);
    RandomLinearNode<Integer> five = new RandomLinearNode<Integer>(5, four, four);
    RandomLinearNode<Integer> six = new RandomLinearNode<Integer>(6, five, two);
    RandomLinearNode<Integer> seven = new RandomLinearNode<Integer>(7, six, three);
    RandomLinearNode<Integer> eight = new RandomLinearNode<Integer>(8, seven, one);

    RandomLinearNode<Integer> newHead = LinkedListUtil.clone(eight);
    assertThat(eight, not(sameInstance(newHead)));
    assertThat(newHead.getElement(), equalTo(eight.getElement()));
    assertThat(newHead.random().getElement(), equalTo(eight.random().getElement()));

    assertThat(newHead.next().getElement(), equalTo(eight.next().getElement()));
    assertThat(newHead.next().random().getElement(), equalTo(eight.next().random().getElement()));

    assertThat(newHead.next().next().getElement(), equalTo(eight.next().next().getElement()));
    assertThat(newHead.next().next().random().getElement(), equalTo(eight.next().next().random().getElement()));


    assertThat(newHead.next().next().next().getElement(), equalTo(eight.next().next().next().getElement()));
    assertThat(newHead.next().next().next().random().getElement(), equalTo(eight.next().next().next().random().getElement()));
}

使用指向除下一节点之外的随机节点的指针复制LinkedList

8 个答案: