Question

我获得了从头开始创建LinkedList的任务，我已经想过如何编写一个在列表末尾添加节点的方法，但我仍然无法想象如何替换节点。以下是我到目前为止的情况：

public boolean replace(int element, int index) {
        Node temp= new Node(pElement);
        Node current = getHead();
        if (!isEmpty()) {
            for (int i = 0; i < index && current.getNext() != null; i++) {
                current = current.getNext();
            }
            temp.setNext(noeudCourant.getNext());
            noeudCourant.setNext(temp);

            listCount++;
            return true;
        }
        return false;
    }

在＆＃34; 0 1 2 3 4 5 6 7 8 9 10＆＃34;

上使用 aNode.replace（10,4）

将进入

[0] - ＆GT; [1] - ＆GT; [2] - ＆GT; [3] - ＆GT; [4] - ＆GT; [ 10 ] - ＆GT; [5] - ＆GT; [6] - ＆GT; [7] - ＆GT; [8] - ＆GT; [9] - ＆GT; [10]

但我想：

[0] - ＆GT; [1] - ＆GT; [2] - ＆GT; [3] - ＆GT; [ 10 ] - ＆GT; [5] - ＆GT; [6] - ＆GT; [7] - ＆GT; [8] - ＆GT; [9] - ＆GT; [10]

感谢任何帮助。

[edit]我已经有了一个工作方法setData（），但我已禁止使用它。我想要的基本上是这样的：

http://i.imgur.com/oOVYCvc.png

Answer 1

如果您真的想要替换节点而不删除它，那么您应该如何编写replace()函数：

public boolean replace(int element, int index) {
    Node head = getHead();
    int counter = 0;
    while(null != head && counter++ < index - 1) {
        head = head.getNext();
    }

    if(null == head || null == head.getNext()) return false;

    Node newNode = new Node(element);
    newNode.setNext(head.getNext().getNext());
    head.setNext(newNode);
    return true;
}

在这里，我假设你的setNext()课程中有Node方法来设置链接。

另请注意，这里假设您永远不会替换head本身，即您永远不会替换索引0处的元素，否则您将不得不返回新的{{1}来自函数。

Answer 2

以下是您问题的简单解决方案：

package linkedlist;

class Node {
    public Node next = null;
    public int element;

    public Node(int el) {
        element = el;
    }
}

class LinkedList {
    public Node first = null;

    public void add(Node node) {
        if (first == null) {
            first = node;
        } else {
            // Traverse to the last
            Node cursor = first;

            while (cursor.next != null) {
                cursor = cursor.next;
            }

            cursor.next = node;
        }
    }

    public void add(int[] elements) {
        int len = elements.length;

        for (int i=0;i < len;i++) {
            add(new Node(elements[i]));
        }
    }

    public boolean replace(int element, int index) {
        Node cursor = first;
        Node prev = null;

        while (cursor != null && index >= 0) {
            index--;
            prev = cursor;
            cursor = cursor.next;
        }

        if (index > 0) return false;

        if (prev != null)
            prev.element = element;

        return true;
    }

    public void displayAll() {
        Node cursor = first;
        while (cursor != null) {
            System.out.print(cursor.element + " ");
            cursor = cursor.next;
        }
        System.out.println();
    }
}

public class Main {

    public static void main(String[] args) {
        // Prepare elements
        LinkedList linkedList = new LinkedList();
        linkedList.add(new int[]{0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10});

        println("Display the initial linked list content:");
        linkedList.displayAll();

        println("After replace:");
        linkedList.replace(10, 4);
        linkedList.displayAll();

        println("Done");
    }

    static void println(String msg) {
        System.out.println(msg);
    }
}

Answer 3

尝试以下代码。无论何时需要，我都在评论中写了逻辑。

public boolean replace(int element, int index) {
    Node temp= new Node(pElement);
    Node current = getHead();
    if (!isEmpty()) {

        //Run for loop one less than index value.
        for (int i = 0; i < index -1 && current.getNext() != null; i++) {
            current = current.getNext();
        }

        // At this point current points to element 3.
        // Set next element of node 4 as a next element of new element 10.

        temp.setNext(current.getNext().getNext());

        // at this point we have two references for element 5 like below
        // [0]->[1]->[2]->[3]->[4]->[5]->[6]->[7]->[8]->[9]->[10]
        // [10]->[5]->[6]->[7]->[8]->[9]->[10]

        // Set null to next element of of element 4 to remove reference to
        // element 5
        current.getNext().setNext(null);

        // At this point we have two list as below:
        // [0]->[1]->[2]->[3]
        // [10]->[5]->[6]->[7]->[8]->[9]->[10]

        // Set new element as next of current element (current element is 3)
        current.setNext(temp);

       // here we have replaced the element 4 with element 10

        listCount++;
        return true;
    }
    return false;
}

LinkedList从头开始，替换节点

3 个答案: