Kotlin中Stack与LinkedList的实现方法示例

class Stack<T : Comparable<T>>(list : MutableList<T>) : Iterator<T> 

// stack的count
 var itCounter: Int = 0

// stack内部实现为MutableList
 var items: MutableList<T> = list

// 判断stack是否为null
fun isEmpty(): Boolean = this.items.isEmpty()

// 获取stack的items counte
fun count(): Int = this.items.count()

// tostring操作
 override fun toString(): String {
  return this.items.toString()
 }

// pop操作，弹出栈顶元素即链表最末端元素，可为null
 fun pop(): T? {
 if (this.isEmpty()) {
  return null
 } else {
  val item = this.items.count() - 1
  return this.items.removeAt(item)
 }
}

// 只读操作，不弹出
fun peek(): T? {
 if (isEmpty()) {
  return null
 } else {
  return this.items[this.items.count() - 1]
 }


}

// hasNext操作
override fun hasNext(): Boolean {
 val hasNext = itCounter < count()
 if (!hasNext) itCounter = 0
 return hasNext
}

// 取next元素
override fun next(): T {
 if (hasNext()){
  val topPos : Int = (count() - 1) - itCounter
  itCounter++
  return this.items[topPos]
 }else{
  throw NoSuchElementException("No such element") // 异常不用new哦
 }
}

class Node<T>(value : T){
 var value : T = value // value可以是任意类型
 var next : Node<T>? = null // next可以为null
 var previous : Node<T>? = null // pre也可以为null
}

// 头结点，引导性作用
var head : Node<T>?= null

// 取决于head是否为null
var isEmpty : Boolean = head == null

// 获取first
fun first() : Node<T>? = head

// 获取last结点，需要一直next才能到达last结点
fun last() : Node<T>?{
 var node = head
 if (node != null){ 
  while (node?.next != null){
   node = node?.next
  }
  return node
 }else{
  return null
 }
}

// 获取count，同样通过next计算
fun count():Int {
 var node = head
 if (node != null){
  var counter = 1
  while (node?.next != null){
   node = node?.next
   counter += 1
  }
  return counter
 } else {
  return 0
 }
}

// append操作,在last结点上append
fun append(value : T){
 var newNode = Node(value)
 // 获取当前节点的最后一个节点
 var lastNode = this.last()
 if (lastNode != null){
  newNode.previous = lastNode
  lastNode.next = newNode
 }else{
  head = newNode
 }
}

// 删除操作
fun removeNode(node : Node<T>) : T{
 val prev = node.previous
 val next = node.next

 if (prev != null){
  prev.next = next
 }else{
  head = next
 }
 next?.previous = prev

 node.previous = null // 将断开的节点前后置null
 node.next = null

 return node.value // 返回删除节点的value
}