Java模拟栈和队列数据结构的基本示例讲解

1、java模拟栈和队列数据结构的基本示例讲解这篇文章主要介绍了 java模拟栈和队列数据结构的基本示例，栈的后进先出和队列的先进先 出是数据结构中最基础的知识，木文则乂对java实现栈和队列结构的方法进行了细分,需要的 朋友可以参考下栈和队列：一般是作为程序员的工具，用于辅助构思算法，生命周期较短，运行时才被创建；访问受限，在特定时刻，只有一个数据可被读取或删除；是一-种抽象的结构，内部的实现机制，对用户不可见，比如用数组、链表來实现栈。模拟栈结构同时，只允许一个数据被访问，后进先出对于入栈和出栈的吋间复朵度都为0(1),即不依赖栈内数据项的个数，操作比较快例，使用数组作为栈的存储结构publi

2、c class stacks<t> private int max; private t ary; private int top; /扌扌旨向栈 顶元素的卜标public stacks(int size) this.max = size; ary = (t) newobjectmax; top = -1;/ 入栈 public void push(t data) 讦(!isfull()ary+top = data; / 出栈 public t pop() 讦(isemptyo) returnnull; return arytop-; / 杳看栈顶 public t peek()

3、returnarytop; 栈是否为空 public boolean isempty() return top =栈是否满 public boolean isfull() return top = max - 1;/size publicint size() return top + 1;public static void main(string args)stacks<lnteger> stack = new stacks<lnteger>(3);for (int i = 0; i < 5; i+)stack.push(i);system.out.printl

4、n("size:" + stack.size(); for (int i = 0;i < 5; i+) integer peek = stack.peek();system.out.println("peek:” + peek);system.out.println("size:" + stack.size(); for (int i = 0; i < 5; i+) integer pop=stack.pop();system.out.pri ntln( "pop:” + pop);system.out.pri ntln

5、("size:" +stack.size();system.out.println("一一")；for (int i = 5; i > 0; i-)stack.push(i);system.out.println(”size:” + stack.size(); for (int i = 5;i > 0; i-) integer peek = stack.peek();system.out.println("peek:" + peek);system.out.println("size:" + stack

6、.size(); for (int i = 5; i > 0; i-) integer pop=stack.pop();system.out.pri ntln ("pop:" + pop);system.out.pri ntln(”size:" +stack.size(); 上面的例了，冇一个maxsize的规定，因为数组是要规定大小的，若想无限制，可以使用其 他结构来做存储，当然也可以new 个新的长度的数组。public stackss() datas =public void push(t data)例，使用linkedlist存储来实现栈public

7、 class stackss<t> private linkedlist<t> datas; new linkedlist<t>(); 入栈datas.addlast(data); 出栈 public t pop() returndatas.removelast(); 查看栈顶 public t peek() returndatas.getlast(); 栈是否为空 public boolean isempty() returndatas.isempty(); /size public int size() return datas.size(); publ

8、ic static void main(string args) stacks<lnteger> stack = new stacks<lnteger(3); for (int i = 0; i < 5; i+) stack.push(i);system.out.pnntln(hsize:h +stack.size(); for (int i = 0; i < 5; i+) integer peek = stack.peek();system.out.pri ntlnpeek:11 + peek);system.out.printlncize:11 + stack

9、.size();for (int i = 0; i < 5; i+) integer pop = stack.pop();system.out.printlnpop:” +pop);system.out.pri ntln(' 'size:*' + stack.size();system.out.println("一一")；for (int i = 5; i > 0; i-) stack.push(i);system.out.println(“size:” + stack.size(); for (int i = 5; i > 0;

10、i-) integer peek=stack.peek();system.out.pri ntln( hpeek:h + peek);system.out.pri ntln(' 'size:" +stack.size(); for (int i = 5; i > 0; i-) integer pop = stack.pop();system.out.printlncpop:1' + pop);system.out.println("size:” + stack.size(); 例,单词逆序,使用statck结构public class word

11、 reverse public static void main(string args) reverse株式会社static void reverse(string word) 讦(word = null) return;stackss<character> stack = new stackss<character>(); word.tochararray();int len = chararrayength;(chararrayi);char chararray = for (int i = 0; i vlen; i+ ) stringbuilder sb = n

12、ewstringbuilder();while ('stack.isemptyo) sb.append(stack.pop();system.out.printin("反转后：11 + sb.tostringo); 打印: 反转后：社会式株模拟队列(一般队列、双端队列、优先级队列)队列：先进先岀，处理类似排队的问题，先排的，先处理，后排的等前血的处理完了，再处理 对于插入和移除操作的时间复杂度都为0(1),从后而插入，从前而移除双端队列：即在队列两端都可以 insert 和 remove： insertleft> insertright, re move left &g

13、t; removeright 含有栈和队列的功能,如去掉insertleft removeleft,那就跟栈一样了；如去掉insertleft> removeright,那就跟队列一样了一般使用频率较低，时间复杂度0(1)优先级队列：内部维护一个按优先级排序的序列。插入时需要比较查找插入的位置，时间复杂度o(n), 删除0(1)/*队列 先进先出，一个指针指示插入的位置，一个指针指示取出数据项的位登*/public class queueqvt> private int max; private t ary; private int front; 队头指名十 指示取出数据项的位置

14、private int rear; /队尾指针指示插入的位置 private int nltems; 实际数据项个数public queueqfint size) this.max = size; ary =仃)newobjectmax; insert(t t) front = 0; rear = -1;nitems = 0; 插入队尾 public voidary+rear = t; n ltems+; 移除队头 aryfront+; if (front = max) /列队到尾了,public t remove) 从头开始t temp =front = 0;nltems-; return

15、temp;查看队头public t peek() return aryfront; public boolean isemptyf) return nitems = 0;public boolean isfull()return nitems = max;public int size() return nitems; public static void main(string args) for (int i = 0; i < 5; i+) queueq<lnteger> queue = new queueq<lnteger>(3);queue.i nsert

16、(i);system.out.pri ntln (“size:” +if (rear = max1) /已到实际队尾，从头开始rear = -1;queue.size(); for (int i = 0; isystem.out.pri ntlnpeek:" + peek);for (int i = 0; i < 5;i+) system.out.println(,remove:m + remove); system.out.pri ntlnr");forsystem.out.pri ntln (''size:11 + queue.size(); pe

17、ek 二 queue.peek();system.out.printl n(”size:" + queue.size(); remove = queue.remove);system.out.pri ntln (“size:" + queue.size();< 5; i+) integer peek = queue.peek();system.out.pri ntln(! 'size:'1 + queue.size();integer remove = queue.remove);system.out.pri ntlnf "size:&quo

18、t; + queue.size();(inti =5; i >0; i-)queue.insert(i);for (inti = 5;i >0;i-) integersystem.out.println(,peek:,+ peek);for (inti = 5;i >0;i-)integersystem.out.pri ntlnremove:”+ remove); /* 双端队列vspan style=l,white-space:pre,> </span>两端插入* 删除 */public classqueueqt<t> private link

19、edust<t> list;public queueqt() list = newlinkedlist<t>(); / 插入队头 public void insertleft(t t) list.addfirst(t); / 插入队尾 public void insertright(t t) list.addlast(t); / 移除队头public t removeleft() return list.removefirst(); / 移除队尾 public tremoveright() return list.removelast(); / 查看队头 public

20、t peekleft() return list.getfirst(); / 杏看队尾 public t peekright() returnlist.getlast(); public boolean isempty() return list.isempty(); public intsize() return list.size(); /*优先级队列 队列中按优先级排序，是一个有序的队列*/public class queueqp private int max; private int ary; private int nitems; 实际数据项个数publicqueueqp(int size) this.max = size; ary = new intmax;nitems =