javascript

Question

我已经使用 javascript 实现了一个单链表。请找到以下代码：

class Node {
  constructor(data) {
    this.data = data;
    this.nextElement = null;
  }
}

class LinkedList {
  constructor() {
    this.head = null;
  }

  isEmpty() {
    return this.head === null;
  }

  insertAtHead(data) {
    const tempNode = new Node(data);
    tempNode.nextElement = this.head;
    this.head = tempNode;
  }

  traverse() {
    let current = this.head;
    while (current.nextElement != null) {
      console.log("node data", current.data);
      current = current.nextElement;
    }
  }

  insertAtTail(data) {
    const tempNode = new Node(data);
    if (this.head === null) {
      this.head = tempNode;
      return;
    }

    let currentNode = this.head;
    while (currentNode.nextElement != null) {
      currentNode = currentNode.nextElement;
    }

    currentNode.nextElement = tempNode;
  }
}

const linkedList = new LinkedList();
linkedList.insertAtTail(12);
linkedList.insertAtTail(23);
linkedList.insertAtTail(25);

linkedList.traverse();

但是遍历方法从不打印最后一个元素。我在这里错过了什么？ insertAtTail 方法看起来是正确的。谁能告诉我一下。

谢谢

Answer 1

当 current.nextElement 为 null 时，

traverse 停止循环——但此时，current 仍然是具有 data 的节点，只是它后面没有 next 节点。

相反，继续前进直到节点本身是 null:

traverse() {
  let current = this.head;
  while (current) { // *** Only change is on this line
    console.log("node data", current.data);
    current = current.nextElement;
  }
}

(那个

while (current) {

可能是

while (current !== null) {

如果你愿意的话。）

实例：

class Node {
  constructor(data) {
    this.data = data;
    this.nextElement = null;
  }
}

class LinkedList {
  constructor() {
    this.head = null;
  }

  isEmpty() {
    return this.head === null;
  }

  insertAtHead(data) {
    const tempNode = new Node(data);
    tempNode.nextElement = this.head;
    this.head = tempNode;
  }

  traverse() {
    let current = this.head;
    while (current) {
      console.log("node data", current.data);
      current = current.nextElement;
    }
  }

  insertAtTail(data) {
    const tempNode = new Node(data);
    if (this.head === null) {
      this.head = tempNode;
      return;
    }

    let currentNode = this.head;
    while (currentNode.nextElement != null) {
      currentNode = currentNode.nextElement;
    }

    currentNode.nextElement = tempNode;
  }
}

const linkedList = new LinkedList();
linkedList.insertAtTail(12);
linkedList.insertAtTail(23);
linkedList.insertAtTail(25);

linkedList.traverse();

Answer 2

在遍历中，需要检查所有节点，直到下一个节点为空。

所以我刚刚从遍历中删除了 .nextElement，它工作正常

class Node {
  constructor(data) {
    this.data = data;
    this.nextElement = null;
  }
}

class LinkedList {
  constructor() {
    this.head = null;
  }

  isEmpty() {
    return this.head === null;
  }

  insertAtHead(data) {
    const tempNode = new Node(data);
    tempNode.nextElement = this.head;
    this.head = tempNode;
  }

  traverse() {
    let current = this.head;
    while (current) { // Here
      console.log("node data", current.data);
      current = current.nextElement;
    }
  }

  insertAtTail(data) {
    const tempNode = new Node(data);
    if (this.head === null) {
      this.head = tempNode;
      return;
    }

    let currentNode = this.head;
    while (currentNode.nextElement != null) {
      currentNode = currentNode.nextElement;
    }

    currentNode.nextElement = tempNode;
  }
}

const linkedList = new LinkedList();
linkedList.insertAtTail(12);
linkedList.insertAtTail(23);
linkedList.insertAtTail(25);


linkedList.traverse();

javascript - 简单的链表遍历问题

javascript - simple linked list traversal issue

traversal

linked-list

data-structures