上海 建站自己建设淘宝客网站需要备案么

数据结构和算法，单链表的实现(kotlin版)

b站视频链接

单链表的实现–koltin版本

1.定义接口，我们需要实现的方法

interface LinkedListAction<E> {fun push(e: E)fun size(): Intfun getValue(index: Int): E?fun insert(index: Int,e: E)fun remove(index: Int)
}

2.定义节点，表示每个链表节点。

data class Node<E>(var next: Node<E>? = null, var value: E)

3.push(e: E)，链表尾部新增一个节点

override fun push(e: E) {val newNode = Node(null, e)if (head != null) {
//            val lastNode = node(len - 1)//O（1）时间复杂度last?.next = newNode} else {head = newNode}last = newNodelen++}

4.size(): Int，返回链表的长度

override fun size(): Int {return len}

5.getValue(index: Int): E?，获取列表的value值

    override fun getValue(index: Int): E? {if (index < 0 || index >= len) {throw ArrayIndexOutOfBoundsException("数组越界.....")}return node(index)?.value}//找到对应index下标的节点。private fun node(index: Int): Node<E>? {var h = head//O（n）时间复杂度for (i in 0 until index) {h = h?.next}return h}

6.insert(index: Int,e: E)，从任意位置插入一个节点

override fun insert(index: Int, e: E) {val newNode = Node(null, e)//考虑边界if (index == 0) {val h = headhead = newNodenewNode.next = h} else {//考虑最后一个位置val prev = node(index - 1)val next = prev?.nextprev?.next = newNodenewNode.next = next}len++}//找到对应index下标的节点。private fun node(index: Int): Node<E>? {var h = head//O（n）时间复杂度for (i in 0 until index) {h = h?.next}return h}

7.remove(index: Int)，任意位置删除一个节点

override fun remove(index: Int) {if (index < 0 || index >= len) {throw ArrayIndexOutOfBoundsException("数组越界.....")}if (index == 0) {val h = headhead = h?.nexth?.next = null} else {val prev = node(index - 1)val current = prev?.nextprev?.next = current?.nextcurrent?.next = null}len--}//找到对应index下标的节点。private fun node(index: Int): Node<E>? {var h = head//O（n）时间复杂度for (i in 0 until index) {h = h?.next}return h}

8.完整Demo

package day1class LinkedList<E> : LinkedListAction<E> {//头指针private var head: Node<E>? = null//优化时间复杂度private var last: Node<E>? = null//集合的长度private var len = 0override fun push(e: E) {val newNode = Node(null, e)if (head != null) {
//            val lastNode = node(len - 1)//O（1）时间复杂度last?.next = newNode} else {head = newNode}last = newNodelen++}//找到对应index下标的节点。private fun node(index: Int): Node<E>? {var h = head//O（n）时间复杂度for (i in 0 until index) {h = h?.next}return h}override fun size(): Int {return len}override fun getValue(index: Int): E? {if (index < 0 || index >= len) {throw ArrayIndexOutOfBoundsException("数组越界.....")}return node(index)?.value}override fun insert(index: Int, e: E) {val newNode = Node(null, e)//考虑边界if (index == 0) {val h = headhead = newNodenewNode.next = h} else {//考虑最后一个位置val prev = node(index - 1)val next = prev?.nextprev?.next = newNodenewNode.next = next}len++}override fun remove(index: Int) {if (index < 0 || index >= len) {throw ArrayIndexOutOfBoundsException("数组越界.....")}if (index == 0) {val h = headhead = h?.nexth?.next = null} else {val prev = node(index - 1)val current = prev?.nextprev?.next = current?.nextcurrent?.next = null}len--}}