Java 如何在单链表中交换两个节点及其内容？

我用谷歌搜索了这个，但他们都在谈论“交换节点而不交换数据”

我尝试自己编写一个交换节点方法：

public void swapNodes(int num1, int num2) {

    if(num1 == num2) {
        return;
    }

    Node currentNum1 = head;
    Node currentNum2 = head;
    Node waitForSwap1 = null;
    Node waitForSwap2 = null;
    while (currentNum1 != null) {
        if (currentNum1.data == num1) {
            waitForSwap1 = currentNum1;
            System.out.println();
            System.out.println("waitForSwap 1");
            System.out.println(waitForSwap1.data);

        }
        currentNum1 = currentNum1.next;
    }

    while (currentNum2 != null) {
        if (currentNum2.data == num2) {
            waitForSwap2 = currentNum2;
            System.out.println("waitForSwap 2");
            System.out.println(waitForSwap2.data);

        }
        currentNum2 = currentNum2.next;
    }

    currentNum1 = waitForSwap2;
    currentNum2 = waitForSwap1;

    System.out.println("currentNum1");
    System.out.println(currentNum1.data);

    System.out.println("currentNum2");
    System.out.println(currentNum2.data);
}

这是结果

如您所见，

currentNum1

和

currentNum2

彼此更改，但打印的结果没有交换。如何交换两个节点及其数据

编辑：下面的完整示例

节点类

public class Node {

public int data;
public Node next;

public Node(int _data) {
    this.data = _data;
    this.next = null;
}

public String toString() {
    return (Integer.toString(data));
}
}

public class LinkedList {

Node head;

public void Insert(int data) {
    Node node = new Node(data);

    if (head == null) {
        head = node;
    } else {
        Node n = head;
        while (n.next != null) {
            n = n.next;
        }
        n.next = node;
    }
}

public void ShowList() {
    Node node = head;
    while (node != null) {
        System.out.print(node.data + " ");
        node = node.next;
    }
}

public void swapNodes(int num1, int num2) {

    if (num1 == num2) {
        return;
    }

    Node currentNum1 = head;
    Node currentNum2 = head;
    Node waitForSwap1 = null;
    Node waitForSwap2 = null;
    while (currentNum1 != null) {
        if (currentNum1.data == num1) {
            waitForSwap1 = currentNum1;
            System.out.println();
            System.out.println("waitForSwap 1");
            System.out.println(waitForSwap1.data);

        }
        currentNum1 = currentNum1.next;
    }

    while (currentNum2 != null) {
        if (currentNum2.data == num2) {
            waitForSwap2 = currentNum2;
            System.out.println("waitForSwap 2");
            System.out.println(waitForSwap2.data);

        }
        currentNum2 = currentNum2.next;
    }

    currentNum1 = waitForSwap2;
    currentNum2 = waitForSwap1;

    System.out.println("currentNum1");
    System.out.println(currentNum1.data);

    System.out.println("currentNum2");
    System.out.println(currentNum2.data);
}
}

链表类

public class Node {

public int data;
public Node next;

public Node(int _data) {
    this.data = _data;
    this.next = null;
}

public String toString() {
    return (Integer.toString(data));
}
}

public class LinkedList {

Node head;

public void Insert(int data) {
    Node node = new Node(data);

    if (head == null) {
        head = node;
    } else {
        Node n = head;
        while (n.next != null) {
            n = n.next;
        }
        n.next = node;
    }
}

public void ShowList() {
    Node node = head;
    while (node != null) {
        System.out.print(node.data + " ");
        node = node.next;
    }
}

public void swapNodes(int num1, int num2) {

    if (num1 == num2) {
        return;
    }

    Node currentNum1 = head;
    Node currentNum2 = head;
    Node waitForSwap1 = null;
    Node waitForSwap2 = null;
    while (currentNum1 != null) {
        if (currentNum1.data == num1) {
            waitForSwap1 = currentNum1;
            System.out.println();
            System.out.println("waitForSwap 1");
            System.out.println(waitForSwap1.data);

        }
        currentNum1 = currentNum1.next;
    }

    while (currentNum2 != null) {
        if (currentNum2.data == num2) {
            waitForSwap2 = currentNum2;
            System.out.println("waitForSwap 2");
            System.out.println(waitForSwap2.data);

        }
        currentNum2 = currentNum2.next;
    }

    currentNum1 = waitForSwap2;
    currentNum2 = waitForSwap1;

    System.out.println("currentNum1");
    System.out.println(currentNum1.data);

    System.out.println("currentNum2");
    System.out.println(currentNum2.data);
}
}

测试仪

public class Runner {

public static void main(String[] args) {
    LinkedList lkdList = new LinkedList();

    lkdList.Insert(10);
    lkdList.Insert(9);
    lkdList.Insert(15);
    lkdList.Insert(2);
    lkdList.Insert(73);

    lkdList.ShowList();

    lkdList.swapNodes(10, 2);

    System.out.println();
    System.out.println("After Swap");

    lkdList.ShowList();
}

}

---

好的，如果您只想交换数据，而不想交换节点，那么如下所示：

  public void swapNodes(int num1, int num2) {

    if (num1 == num2) {
      return;
    }

    Node node1 = null;
    Node node2 = null;

    Node cur = head;
    while(cur != null) {
      if (num1 == cur.data) {
        node1 = cur;
      }
      if (num2 == cur.data) {
        node2 = cur;
      }
      cur = cur.next;
    }

    if (node1 == null || node2 == null)
      return;

    int tmp = node1.data;
    node1.data = node2.data;
    node2.data = tmp;
  }

---

好的，如果您只想交换数据，而不想交换节点，那么如下所示：

  public void swapNodes(int num1, int num2) {

    if (num1 == num2) {
      return;
    }

    Node node1 = null;
    Node node2 = null;

    Node cur = head;
    while(cur != null) {
      if (num1 == cur.data) {
        node1 = cur;
      }
      if (num2 == cur.data) {
        node2 = cur;
      }
      cur = cur.next;
    }

    if (node1 == null || node2 == null)
      return;

    int tmp = node1.data;
    node1.data = node2.data;
    node2.data = tmp;
  }

---

请拿出，这样我们就可以简单地复制粘贴您的代码并进行调试。另外，为什么“交换节点而不交换数据”对您不起作用？您实际上不是交换节点，您只是将找到的节点分配给节点对象并打印它们的值。为了交换，您还需要跟踪上一个节点，因为要交换，您还将更改previous.next的值；您只需更改局部变量引用的内容。@SergeiSirik这是作业的要求，教授希望我们同时更改节点和数据。因此，如果我没有弄错，这正是您要寻找的“交换节点而不交换数据”-这意味着节点对象将被交换，或者您有两个不同的作业：1。交换节点；2在不交换节点的情况下交换节点中的数据？请拿出，这样我们就可以简单地复制粘贴您的代码并进行调试。另外，为什么“交换节点而不交换数据”对您不起作用？您实际上不是交换节点，您只是将找到的节点分配给节点对象并打印它们的值。为了交换，您还需要跟踪上一个节点，因为要交换，您还将更改previous.next的值；您只需更改局部变量引用的内容。@SergeiSirik这是作业的要求，教授希望我们同时更改节点和数据。因此，如果我没有弄错，这正是您要寻找的“交换节点而不交换数据”-这意味着节点对象将被交换，或者您有两个不同的作业：1。交换节点；2是否在不交换节点的情况下交换节点中的数据？