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javascript - javascript - 简单的链表遍历问题

问题描述

我已经使用 javascript 实现了一个单链表。请在下面找到代码：

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 traverse 方法从不打印最后一个元素。我在这里想念什么？虽然 insertAtTail 方法看起来是正确的。谁能告诉我。

谢谢

标签： javascriptdata-structureslinked-listtraversal

解决方案

traversecurrent.nextElement当is时停止它的循环null ——但在那一点上，current它仍然是一个带有 的节点data，只是它后面没有下一个节点。

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

while (current) {

可能

while (current !== null) {

如果你更喜欢。）

现场示例：

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();

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