JS实现单链表

单链表的操作核心有：

push（value） - 在链表的末尾/头部添加一个节点

pop（） - 从链表的末尾/头部删除一个节点

get（index） - 返回指定索引处的节点

delete（index） - 删除指定索引处的节点

isEmpty（） - 根据列表长度返回true或false

print（） - 返回链表的可见表示

实现如下：

class Node {
  constructor(data) {
    this.data = data
    this.next = null
  }
}
class LinkedList {
  constructor() {
    this.head = null
    this.tail = null
    this.length = 0
  }
  push (data) {
    const node = new Node(data) // 创建一个新节点
    if (this.head === null) { // 检查头部是否为空
      this.head = node
      this.tail = node
    }
    this.tail.next = node
    this.tail = node
    this.length++
  }
  pop () {
    if (this.isEmpty()) {
      return null
    }
    if (this.head === this.tail) {
      this.head = null
      this.tail = null
      this.length--
      return this.tail
    }
    let node = this.tail
    let currentNode = this.head
    let penultimate
    while (currentNode) {
      if (currentNode.next === this.tail) {
        penultimate = currentNode
        break
      }
      currentNode = currentNode.next
    }
    penultimate.next = null
    this.tail = penultimate
    this.length--
    return node
  }
  get (index) { // 处理边界条件
    if (index === 0) {
      return this.head
    }
    if (index < 0 || index > this.length) {
      return null
    }
    let currentNode = this.head
    let i = 0
    while (i < index) {
      i++
      currentNode = currentNode.next
    }
    return currentNode
  }
  delete (index) {
    let currentNode = this.head
    if (index === 0) {
      let deleteNode
      currentNode.next = this.head
      deleteNode = currentNode
      this.length--
      return deleteNode
    }
    if (index < 0 || index > length) {
      return null
    }
    let i = 0
    let previous
    while (i < index) {
      i++
      previous = currentNode
      currentNode = currentNode.next
    }
    previous.next = currentNode.next
    this.length--
    return currentNode
  }
  isEmpty () {
    return this.length === 0
  }
  print () {
    const list = []
    let currentNode = this.head
    while (currentNode) {
      list.push(currentNode.data)
      currentNode = currentNode.next
    }
    return list.join('=>')
  }
}

// test
const l = new LinkedList()
const valuesB = ['A', 'B', 'C', 'D', 'E']
valuesB.forEach(value => l.push(value))