实现整数集合的并、交、差运算

  问题:编写一个程序,实现一个整数集合的基本运算
  要求:输入整数集合{2,4,1,3,5}和{2,5,10},输出前者的元素个数以及它们进行集合的并、交、差运算
   设计一个类IntSet,包括私有数据成员len(集合的长度)和数组s(存放集合元素),以及如下方法:
    public void insert(int d)    向集合中添加一个元素,重复的元素不能添加
    public int length()      返回集合的元素个数
    public int getInt(int i)             返回集合中位置i的元素
    public void disp()      输出集合的所有元素
    public IntSet union(IntSet s2)  实现两个集合的并运算
    public IntSet intersection(IntSet s2)  实现两个集合的交运算
    public IntSet difference(IntSet s2)  实现两个集合的差运算
    public IntSet copySet(IntSet s2)   实现两个集合的拷贝
 
Java代码:
public class IntSet {
       private int len;  //集合长度
       private int[] s;  //数组
       
       public IntSet() {
              len = 0;
              s = new int[100];
       }
    public void resize()  //重置数组
    {
        int[] b = new int[2*s.length];
        System.arraycopy(s, 0, b, 0, s.length);
        s = b;    
    }
       public void insert(int d) //集合中插入元素
       {
             if(len>=s.length)
             {
                 resize();
              }
              for(int i=0;i<len;i++) {
                     if(s[i]==d){
                           return;
                     }
              }
          s[len]=d;
          len++;
       }
       
       public int length() //获取集合长度
       {
              return len;
       }
       
       public int getInt(int i) {
              
              if(i>=0 && i<len)
              {
                     return s[i];
              }else {
                     return -1;
              }
       }
       
       public void disp() {
              for(int i=0;i<len;i++)
              {
                     System.out.print("\t"+s[i]);
              }
       }
       
       public IntSet union(IntSet s2) //并
       {
              int same;
              for(int i=0;i<this.len;i++)
              {
                     same = 0;
                     for(int j=0;j<s2.len;j++)
                     {
                           if(this.getInt(i)==s2.getInt(j))
                           {
                                  same = 1;
                                  break;
                           }
                           if(same == 0)
                           {
                                  this.insert(s2.getInt(j));
                           }
                     }
              }
              return this;
       }
       
       public IntSet intersection(IntSet s2) { //交
              
              int same;
              for(int i=0;i<this.len;i++) {
                     same = 0;
                     for(int j=0;j<s2.len;j++) {
                           if(this.s[i] == s2.getInt(j))
                           {
                                  same = 1;  //找到重复的元素,则same=1
                                  break;
                           }
                     }
                           if(same == 0) {//找到不重复的元素,则删除
                                  for(int k=i;k<this.len;k++)
                                  {
                                         this.s[k] =this.s[k+1];
                                  }
                                  
                                  i--;
                                  this.len--;
                                  
                           }
                     
              }
              return this;
       }
       public IntSet difference(IntSet s2) { //差
              
              int same;
              for(int i=0;i<this.len;i++) {
                     same = 0;
                     for(int j=0;j<s2.len;j++) {
                           if(this.s[i] == s2.getInt(j))
                           {
                                  same = 1; //找到重复的元素,则same=1
                                  break;
                           }
                     }
                           if(same == 1) {//找到重复的元素,则删除
                                  for(int k=i;k<this.len;k++)
                                  {
                                         this.s[k] =this.s[k+1];
                                  }
                                  
                                  i--;
                                  this.len--;
                                  
                           }
                     
              }
              return this;
       }
       
       public IntSet copySet(IntSet s2) {
              this.len = s2.len;
              for(int i=0;i<s2.len;i++) {
                     this.s[i] = s2.getInt(i);
              }
              return this;
       }
}
 
测试代码:
  public class Test {
       public static void main(String[] args) {
              IntSet s1,s2;
              s1 = new IntSet();
              s2 = new IntSet();
              s1.insert(2);
              s1.insert(4);
              s1.insert(1);
              s1.insert(3);
              s1.insert(5);
              
              s1.disp();
              
              s2.insert(2);
              s2.insert(5);
              s2.insert(10);
              
              System.out.println();
              s2.disp();
              
//            s1.union(s2);
//            System.out.println();
//            s1.disp();
              
//            s1.intersection(s2);
//            System.out.println();
//            s1.disp();
              
//            s1.difference(s2);
//            System.out.println();
//            s1.disp();
              
//            IntSet s6 = new IntSet();
//            s6.copySet(s2);
//            s6.disp();
              
              s2.difference(s1);
              System.out.println();
              s2.disp();
       }
}
 
posted @   #独狼  阅读(1248)  评论(0编辑  收藏  举报
编辑推荐:
· AI与.NET技术实操系列:基于图像分类模型对图像进行分类
· go语言实现终端里的倒计时
· 如何编写易于单元测试的代码
· 10年+ .NET Coder 心语,封装的思维:从隐藏、稳定开始理解其本质意义
· .NET Core 中如何实现缓存的预热?
阅读排行:
· 分享一个免费、快速、无限量使用的满血 DeepSeek R1 模型,支持深度思考和联网搜索!
· 基于 Docker 搭建 FRP 内网穿透开源项目(很简单哒)
· 25岁的心里话
· ollama系列01:轻松3步本地部署deepseek,普通电脑可用
· 按钮权限的设计及实现
点击右上角即可分享
微信分享提示