[数据结构与算法]leetcode(2链表）

1.单向链表

class Node{
    constructor(element,next){
        this.element = element
        this.next = next
        }
}


class LinkedList{
    constructor(){
        this.head = null;
        this.size = 0;
    }
    add(index,element){
        if(arguments.length === 1){
            element = index
            index = this.size
        }
        if(index < 0 || index > this.size) throw new Error('越界')
        if(index === 0 ){
            // 取到老的头
            let head = this.head
            this.head = new Node(element,head)
        }else{
                // 找到当前索引的上一个
                let prevNode = this.getNode(index -1)
                prevNode.next = new Noe(element,prevNode.next)
            }
        this.size++
    }
    getNode(index){
        let current = this.head;
        for(let i =0; i< index;i++){
            current = current.next
        }
        return current
    }
}
const ll = new LinkedList()
ll.add(0,100)
ll.add(0,200)
ll.add(300)
ll.add(1,500)
console.dir(ll,{depth:1000})