ADT基础(一)—— List,Stack,and Queue
ADT基础(一)—— List,Stack,and Queue
1 List 表示
数组:易于search,难于insert和remove
链表:难于search,易于insert和remove
//Node类,LinkedList类
public class LinkedList{
Node head = null;
class Node{ //element和next
object element;
Node next;
Node(object e){
this.element = e;
}
}
//输出链表并获取长度
int getLength(){
int length = 0;
Node tmp = head;
while(tmp != null){
length++;
System.out.println(tmp.data);
tmp = tmp.next;
}
return length;
}
//查询element为e的第一个位置
int getIndex(object e){
int index = -1;
Node tmp = head;
while(tmp!=null){
index++:
if(tmp.element == e){
return index;
}
tmp = tmp.next;
}
return -1;
}
//获取指定位置的element
object getObject(int index){
if(index<0||index >= getLength()) //print fault;
Node tmp = head;
if(head==null) //print fault;
for(int i=0;i<index;i++){
tmp = tmp.next;
}
return tmp.element;
}
//头插法
void addHead(object e){
Node newNode = new Node(e);
newNode.next = head;
head = newNode;
}
//尾插法
void addTail(object e){
Node newNode = new Node(e);
if(head == null) head = newNode;
else{
Node tmp = head;
while(tmp.next != null){
tmp = tmp.next;
}
tmp.next = newNode;
}
}
//随机节点插入法
void insert(int index,object e){
int size = getLength();
if(index>=size||index<0) //print fault;
if(index==0) addHead(e);
else if(index == size-1) addTail(e);
else{
Node pre = head;
Node cur = head.next;
for(int i=0;i<index-1;i++){
pre = pre.next;
cur = cur.next;
}
//pre保存索引上一个节点,cur保存索引值当前节点
Node newNode = new Node(e);
pre.next = newNode;
newNode.next = cur;
}
}
//删除头节点
void deleteHead(){
if(head == null) return;
head = head.next;
}
//删除尾节点
void deleteTail(){
if(head==null) return;
Node btmp = head;
Node tmp = btmp.next;
if(tmp == null){
head = null;
return;
}
while(tmp.next != null){
btmp = tmp;
tmp = tmp.next;
}
btmp.next = null;
}
//随机删除节点
void remove(int index){
int size = getLength();
if(index<0||index>=size) //print fault;
if(index == 0) deleteHead();
else if(index == size-1) deleteTail();
else{
Node pre = head;
for(int i=0;i<index-1;i++){
pre = pre.next;
}
pre.next = pre.next.next;
}
}
}
//由单链表的增加删除可以看出,链表想要对指定索引进行操作(增加,删除),则必须获取该索引的前一个元素。记住这句话,对链表算法题很有用。
2 Stack表示
- 后入先出
//栈的链表实现,栈顶在topOfStack,即head处;
//push和pop都在head处
public class StackLi
{
public StackLi( ){ topOfStack = null; }
public boolean isFull( ){ return false; }
public boolean isEmpty( ){ return topOfStack = = null; }
public void makeEmpty( ){ topOfStack = null; }
public void push( object x){
topOfStack = new ListNode(x,topOfStack);
}
public object top(){
if(topOfStack == null) return null;
return topOfStack.element;
}
public void pop() throws Underflow{
if(topOfStack == null) throw new Underflow();
topOfStack = topOfStack.next;
}
public object topAndPop( ){
if(topOfStack == null) return null;
object res = topOfStack.element;
topOfStack = topOfStack.next;
return res;
}
private ListNode topOfStack;
}
//栈的数组实现,栈顶在topOfStack,即在数组n-1位上(假设压入n个元素);
//push和pop依次向后或向前
public class stackAr{
public StackAr( ){
this(DEFAULT_CAPACITY);
}
public StackAr(int capacity){
theArray = new object[capacity];
topOfStack = -1;
}
public boolean isEmpty( ){ return topOfStack == -1; }
public boolean isFull( ){ return topOfStack == theArray.length –1; }
public void makeEmpty( ){ topOfStack = -1; }
public void push( object x ) throws overflow{
if(topOfStack == theArray.length - 1) throw new Overflow();
topOfStack++;
theArray[topOfstack] = x;
}
public object top( ){
if(topOfStack==-1) return null;
return theArray[topOfStack];
}
public void pop( ) throws Underflow{
if(topOfStack == -1) throw new Undewflow();
theArray[topOfStack] == null;
topOfStack--;
}
public object topAndPop( ){
if(topOfStack==-1) return null;
object res = theArray[topOfStack];
theArray[topOfStack] == null;
topOfStack--;
return res;
}
private object [ ] theArray;
private int topOfStack;
static final int DEFAULT_CAPACITY = 10;
}
3 Queue表示
- 插入与删除在不同端,先入先出
//队列的数组实现,在front位删除,在back位插入 front到back由0到n-1
public class QueueAr
{
public QueueAr(){
this(DEFAULT_CAPACITY);
}
public QueueAr( int capacity){
theArray = new Object[capacity];
currentSize = 0;
front = 0;
back = -1;
}
public boolean isEmpty( ){ return currentsize == 0; }
public boolean isfull( ){ return currentSize == theArray.length; }
public Object getfront( )
public void enqueue( Object x ) throw Overflow{
if(currentSize == theArray.length) throw new Overflow();
back++;
if(back == theArray.length) back = 0; //队列满则新元素回到0位插入
theArray[back] = x;
currentSize++;
}
private Object dequeue( ){
if(currentSize == 0) return null;
curretSize--;
object res = theArray[front];
theArray[front] = null;
front++;
if(front == theArray.length) front = 0; //队列删到尾则回到0删除
return res;
}
private Object [ ] theArray;
private int currentSize;
private int front; //删除
private int back; //插入
static final int DEFAULT_CAPACITY = 10;
}
//队列的链表实现,front在head
public class LinkedQueue
{
public LinkedQueue(){
this.head = null;
this.tail = null;
this.size = 0;
}
public boolean IsEmpty(){return size==0;}
public boolean IsFull(){return false};
public void add(object x){
if(size==0){
head = new Node(x);
tail = head;
size++;
}else{
tail.next = new Node(x);
tail = tail.next;
size++;
}
}
public object delete(){
if(size==0) return null;
object res = head.element;
head = head.next;
if(head == null) tail = null; //head为null,代表已经行进到tail.next,此时为空链表
size--;
return res;
}
private Node head;
private Node tail;
private int size;
};
ADT基础(一)—— List,Stack,and Queue的更多相关文章
- page74-泛型可迭代的基础集合数据类型的API-Bag+Queue+Stack
[泛型可迭代的基础集合数据类型的API] 背包:就是一种不支持从中删除元素的集合数据类型——它的目的就是帮助用例收集元素并迭代遍历所有收集到的元素.(用例也可以检查背包是否为空, 或者获取背包中元素的 ...
- bfs和dfs辨析—基础复习(从stack和queue的角度来理解区别,加深理解,不再模糊)
参考: https://www.cnblogs.com/Tovi/articles/6194815.html https://blog.csdn.net/dangzhangjing97/article ...
- Java容器:Stack,Queue,PriorityQueue和BlockingQueue
Stack Queue PriorityQueue BlockingQueue ArrayBlockingQueue LinkedBlockingQueue PriorityBlockingQueue ...
- ADT基础(二)—— Tree,Heap and Graph
ADT基础(二)-- Tree,Heap and Graph 1 Tree(二叉树) 先根遍历 (若二叉树为空,则退出,否则进行下面操作) 访问根节点 先根遍历左子树 先根遍历右子树 退出 访问顺序为 ...
- Java集合的Stack、Queue、Map的遍历
Java集合的Stack.Queue.Map的遍历 在集合操作中,常常离不开对集合的遍历,对集合遍历一般来说一个foreach就搞定了,但是,对于Stack.Queue.Map类型的遍历,还是有一 ...
- [STL]deque和stack、queue
怎么说呢,deque是一种双向开口的连续线性空间,至少逻辑上看上去是这样.然而事实上却没有那么简单,准确来说deque其实是一种分段连续空间,因此其实现以及各种操作比vector复杂的多. 一.deq ...
- python实现之极简stack和queue
用python实现一个极简的stack和queue,那是so easy的事情了,简洁易懂,适合小白~ 直接上代码吧: node: class LinkNode: def __init__( self, ...
- C++ STL stack和queue
C++ STL中独立的序列式容器只有vector,list,deque三种,stack和queue其实就是使用容器适配器对deque进行了封装,使用了新接口. 使用标准库的栈和队列时,先包含相关的头文 ...
- 剑指offer——stack与queue的互相实现
我们知道,stack和queue是C++中常见的container.下面,我们来探究下如何以stack来实现queue,以及如何用queue来实现stack. 首先,先了解下stack与queue的基 ...
随机推荐
- 翻译:《实用的Python编程》00_Setup
课程设置与概述 欢迎访问本课程(Practical Python Programming).这个页面包含一些关于课程设置的重要信息. 课程周期和时间要求 该课程最初是作为一个由讲师主导的,持续 3 - ...
- Educational Codeforces Round 88 (Rated for Div. 2) D. Yet Another Yet Another Task(枚举/最大连续子序列)
题目链接:https://codeforces.com/contest/1359/problem/D 题意 有一个大小为 $n$ 的数组,可以选取一段连续区间去掉其中的最大值求和,问求和的最大值为多少 ...
- AtCoder Beginner Contest 181 E - Transformable Teacher (贪心,二分)
题意:有一长度为奇数\(n\)的数组\(a\),和长度为\(m\)的数组\(b\),现要求从\(b\)中选择一个数放到\(a\)中,并将\(a\)分成\((n+1)/2\)个数对,求最小的所有数对差的 ...
- 一篇文章图文并茂地带你轻松学完 JavaScript 设计模式(二)
JavaScript 设计模式(二) 本篇文章是 JavaScript 设计模式的第二篇文章,如果没有看过我上篇文章的读者,可以先看完 上篇文章 后再看这篇文章,当然两篇文章并没有过多的依赖性. 5. ...
- Error Code: 1055.Expression #1 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'userinfo.
环境:mysql-8.0.15-winx64 问题描述: Error querying database. Cause: java.sql.SQLSyntaxErrorException: Expre ...
- Gym 101480I Ice Igloos(思维乱搞)题解
题意:给个最多500 * 500的平面,有半径最多不为1的n个圆,现在给你1e5条线段,问你每条线段和几个圆相交,时限10s 思路: 因为半径<1,那么我其实搜索的范围只要在线段附近就好了.x1 ...
- Ubuntu16安装chrome
不免让您失望, 安装正常的chrome,Dependency is not satisfiable: libnss3 (>= 2:3.22)问题一直没能解决,故使用chromium次而代之. s ...
- iPhone 12 Pro 屏幕时间设置的密码锁出现弹窗 UI 错位重大 Bug
iPhone 12 Pro 屏幕时间设置的密码锁出现弹窗 UI 错位重大 Bug iOS 14.1 Bug 弹窗 UI 非常丑 弹窗屏占太高了 屏幕使用时间 https://support.apple ...
- React Hooks: useContext All In One
React Hooks: useContext All In One useContext https://reactjs.org/docs/hooks-reference.html#useconte ...
- flex & flex-wrap
flex & flex-wrap https://css-tricks.com/almanac/properties/f/flex-wrap/ https://developer.mozill ...