链表题目

https://leetcode-cn.com/problemset/algorithms/?topicSlugs=linked-list

首先基本上会把arr=[1,2,3] ->转为1->2->3 这样的结构

class Node {
  constructor(item) {
    this.item = item
    this.next = null

  }
}
class linkedList {
  constructor() {
    this.head = null
    this.length = 0

  }
  append(item) {
    const node = new Node(item)
    let current = this.head
    let previous = null
    if (!current) this.head = node
    else {
      while (current) {
        previous = current
        current = current.next
      }
      previous.next = node

    }
    this.length++
  }
  toString() {
    const arr = []
    let current = this.head
    if (!current) return ''
    while (current) {
      arr.push(current.item)
      current = current.next

    }
    return arr.join('->')
  }
}

View Code

const arr = [1, 1, 2]
const link = new linkedList()
arr.forEach(item => link.append(item))
console.log(link.toString()); //1->1->2

至此就完成了数组转链表的操作

使用：link.head = deleteDuplicates(link.head) deleteDuplicates是一个举例子就是操作函数了

删除链表中的节点

function deleteNode(node) {
  let current = link.head
  let previous = null
  while (current) {
    if (current.item === node) break
    previous = current
    current = current.next
  }
  previous.next = current.next
  console.log(link.toString());


}
deleteNode(node)

View Code