实验三-栈和队列

1、实验报告三 栈和队列一、 实验目的：1掌握栈的根本操作的实现方法。2利用栈先进后出的特点，解决一些实际问题。3掌握链式队列及循环队列的根本操作算法。4应用队列先进先出的特点，解决一些实际问题。 二、 实验内容：1、 使用一个栈，将一个十进制转换成二进制。 粘贴源程序： package Q1;public class SeqStack publicintelement ;publicinttop ;publicstaticSeqStack p;publicSeqStack( int size)this . element =new int size; this . top =-1;public

2、void push( int x)this . top +;this . element this . top =x;public int pop(). top -;. top ;return this . top =-1 ? -1: (int )this . element thispublic int get()return this . top =-1 ? -1: (int )this . element thispublic static void disp(SeqStack p)int t = -2;while (t!=-1) t=p.pop(); if (t!=-1)System.

3、 out .printf( "%d" ,t);public static void fun( int x)int t;while (x!=1)t=x%2;x=x/2; p.push(t);if (x=1)p.push(x);public static void main(String args) p=new SeqStack(13);fun (99);disp ( p);粘贴测试数据及运行结果：<terminated110&011|2、回文是指正读反读均相同的字符序列，如"acdca、“dceecd 均是回文，但"book不是回文。利用1中

4、的根本算法，试写一个算法判定给定的字符串是否为回文。提示：将半字符入栈，依次弹出与另一半逐个比拟 粘贴源程序:package Q2;public classSeqStack publicint element ;publicint top ;publicstatic SeqStackp；publicSeqStack( intsize)this.element =newint size;this.top =-1;public void push( int x)this . top +;this . element this . top =x;public int pop()return this

5、 . top =-1 ? -1:(public int get()return this . top =-1 ? -1:(int ) this . element this . top -;int ) this . element this . top ;public static void input(String str)int i=0;int t=str.le ngth();if (t%2=0)for (i=0;i<t/2;i+)p.push(str.charAt(i); else for (i=0;i<=t/2;i+)p.push(str.charAt(i);public

6、staticboolea n compare(Str ing str,SeqStack p)boolean flag = true ;char t;int len gth=str.le ngth();if (length%2=0)for (int i=0;i<length/2;i+)t=str.charAt(le ngth/2+i);if (t!=p.pop()flag= false ; break ; else for (int i=0;i<length/2+1;i+)t=str.charAt(le ngth/2+i);if (t!=p.pop()flag= false ; br

7、eak ;return flag;public static void main( Str in g args) boolea n flag;p=new SeqStack(100);Stri ng str = new String( "acbca");p. input (str):flag= p pare (str. p):if (flag= true )System. out .println("yes");else System. out .println( "No");粘贴测试数据及运行结果：3、使用3个队列分别保存上最近10个

8、“未接来电、“已接来电、“已拨粘贴源程序:package Q3;public class SeqQueue<T> publicObject element ;publicint front,rear ;publicSeqQueuep；publicSeqQueue(int len gth)this.element=new Objectle ngth;this.front =this . rear =0;public boolea n isEmpty()return this . front = this . rearpublicboolea n isFull()boolea n fl

9、ag= false ;if (this .front != this . rear && ( this . rear +1)%this . eleme nt this . front )flag= true ;return flag;public void enquere(T x)this . element this . rear =x;this . rear =( this . rear +1)%this . element . length ;public T dequeue()if (isEmpty()return n ull ;T temp=(T) this _ele

10、ment Lthis .front 丄this . front =( this . front +1)%this . element . length ;return temp;public static void disp(SeqQueue p)int i=1;.length =while (p.isEmpty()!= true )System. out .println(i+"."+p.dequeue();i+；System. out .println();public static void fun() long num=0;new SeqQueue(11);new

11、SeqQueue(11);int t=0;SeqQueue missedCall =SeqQueue receivedCall =SeqQueue outgo in gCall =new SeqQueue(11);while (!(missedCall.isFull() && receivedCall.isFull() && outgoi ngCall.isFull()num=( int )(Math.random ()*100000000);t=( int )(Math. random ()*101);if (t%3=0)missedCall.e nquere

12、 (n um);if (t%3=1)receivedCall.e nquere( num);if (t%3=2)outgo in gCall.e nquere (n um);"missedCall" );"receivedCall")"outgoi ngCall")main( Stri ng args) System. out .println( disp (missedCall);System. out .println( disp (receivedCall);System. out .println( disp (outgoingCall);public static voidfun ();粘贴测试数据及运行结果:4.41ME2E百用7W27A6127S82631824«57rsceiyedcall&47S*,4697426.7783 &05 37.92174S25均眈弼旳outgpi ngCall1-B324BM72.m3tl9353.274277285u673107S56-777&&