• 数据结构--开放定址法解决散列冲突时几种探测法的比较


         开放定址法解决散列冲突时主要有线性探测法,平方探测法和双散列法,以下代码通过插入大量随机数,来统计几种探测法产生冲突的次数。

        

    #include<iostream>
    
    using namespace std;
    
    #define MinTablesize 10
    #define Num 3000
    typedef unsigned int Index;
    typedef Index Position;
    
    
    
    struct Hashtal;
    typedef struct Hashtal *Hashtable;
     
    int num_quadratic_probing = 0;
    int num_liner_probing = 0;
    int num_double_probing = 0;
    
    int array_rand[Num];    //大小为1000的随机数组
    
    enum KindOfEntry  //枚举类型,只能是其中的一个值
    {
      Legitimate,Empty,Delete
    };
     
    struct HashEntry
    {
       int Element;
       enum KindOfEntry Info;
    };
     
    typedef struct HashEntry Cell;
     
    struct Hashtal
    {
      int TableSize;
      int Insert_num;   //记录插入的个数
      Cell *Thecells;
    };
    
     Hashtable Rehash (Hashtable H,int ty);
    Hashtable Insert(int Key,Hashtable H,int ty);
    
    int Hash (int x,Hashtable H)   //散列函数
    {
       return x % H->TableSize;
    }
     
    int NextPrime (int Key)  // 寻找Key后的第一个素数
    {
        if(Key == 1 || Key == 2)
            return Key;
     
        bool flag = 0;
        int pre_Key;
        while(!flag)
        {
           pre_Key = Key;
           for(int i = 2; i != Key; ++i)
           {
              if(Key % i == 0)   //Key不是素数
              {
                 Key++;
                 break;
              }       
           }
           if(pre_Key == Key)  //说明Key没有改变
               flag = 1;
           else
               flag = 0;
     
        }
        return Key;
    }
    Hashtable InitTable(int Tablesize)   //创建一个散列表
    {
       Hashtable H;
       if(Tablesize < MinTablesize)
       {
         cout << "Table is too small" << endl;
         return NULL;
       }
     
       //创建散列表
       H = (Hashtable)malloc(sizeof(struct Hashtal));
       if(H == NULL)  cout << "out of space " << endl;
       H->TableSize = NextPrime(Tablesize);  //大小为H->TableSize后的第一个素数
       H->Thecells = (Cell*)malloc(sizeof(Cell) * H->TableSize );
       if(H->Thecells == NULL)  cout << "out of space " << endl;
       H->Insert_num = 0;
       for (int i = 0; i != H->TableSize; ++i)
          H->Thecells[i].Info = Empty;   //每个元素都标记为空
       return H;
    }
    
    Position Find_linear_probing (int Key , Hashtable H)   //使用线性探测散列
    {
       Position CurrPos;
       int CollNum;
       CollNum = 0;
       CurrPos = Hash(Key,H);
       
       while(H->Thecells[CurrPos].Info != Empty && H->Thecells[CurrPos].Element != Key)  //找不到时返回一个空单元
       {
          CurrPos =  (CurrPos + (++CollNum)) % H->TableSize;    //使用线性探测散列法
    	  ++num_liner_probing;     //冲突次数加1
          if((int)CurrPos >= H->TableSize )
              CurrPos -= H->TableSize;
       }
       return CurrPos;
    }
    
    
    
    Position Find_quadratic_probing (int Key , Hashtable H)   //使用平方探测散列
    {
       Position CurrPos;
       int CollNum;
       CollNum = 0;
       CurrPos = Hash(Key,H);
       while(H->Thecells[CurrPos].Info != Empty && H->Thecells[CurrPos].Element != Key)  //找不到时返回一个空单元
       {
    	   ++CollNum;
          CurrPos =  (CurrPos + CollNum * CollNum) % H->TableSize;    //使用平方探测散列法
    	  ++num_quadratic_probing;      //冲突次数加1
          if((int)CurrPos >= H->TableSize )
              CurrPos -= H->TableSize;
       }
       return CurrPos;
    }
    
    Position Find_double (int Key , Hashtable H)   //使用双散列
    {
       Position CurrPos;
       int CollNum;
       CollNum = 0;
       CurrPos = Hash(Key,H);
       while(H->Thecells[CurrPos].Info != Empty && H->Thecells[CurrPos].Element != Key)  //找不到时返回一个空单元
       {
          CurrPos = (CurrPos + (++CollNum) * (7 - (Key % 7 ))) % H->TableSize;    //使用双散列探测
    	  ++num_double_probing;      //冲突次数加1
          if((int)CurrPos >= H->TableSize )
              CurrPos -= H->TableSize;
       }
       return CurrPos;
    }
    
    Hashtable Insert(int Key,Hashtable H,int ty)
    {
      if(((double)H->Insert_num /H->TableSize) > 0.5)  //如果装填因子大于0.5,就再散列,
    	  H = Rehash(H,ty);    
    
      Position Pos;
      switch (ty)
      {
         case 1: 
    		 Pos = Find_quadratic_probing(Key,H);
    		 break;
         case 2: 
    		 Pos = Find_linear_probing(Key,H);
    		 break;
         case 3: 
    		 Pos = Find_double(Key,H);
    		 break;
    	 default:
    		 cout << "Insert type error" << endl;
      }
      if(H->Thecells[Pos].Info != Legitimate)  //
      {
         H->Thecells[Pos].Element  = Key;
         H->Thecells[Pos].Info = Legitimate;
    	 H->Insert_num++;    //插入的元素个数加1
      }
      return H;
    }
     Hashtable Rehash (Hashtable H,int ty)   //再散列,把表的大小放大到原来的2倍,再把原来的元素插入到新散列表中
    {
        int Oldsize;
        Cell *Oldcells;
     
        Oldcells = H->Thecells;
        Oldsize = H->TableSize;
     
        H = InitTable(2 * Oldsize);
     
        for (int i = 0; i != Oldsize; ++i)
        {
          if(Oldcells[i].Info == Legitimate)
              Insert(Oldcells[i].Element, H,ty);
        }
        free(Oldcells);
        return H;
    }
    
    void destroyTable(Hashtable H) 
    {  
        free(H->Thecells );  
        free(H);  
    } 
    
    
    
    int main ()
    {
      
      for (int i = 0; i != Num; ++i)
      {
         array_rand[i] = rand();  //产生随机数
      }
    
    
      Hashtable H1 = InitTable(200);  //200大小的散列表
    
      for(int i = 0; i != Num; ++i)
      {
        H1 = Insert(array_rand[i],H1,1);
      }
      cout << num_quadratic_probing << endl;
      destroyTable(H1);
    
      Hashtable H2 = InitTable(200);  //200大小的散列表
      for(int i = 0; i != Num; ++i)
      {
        H2 = Insert(array_rand[i],H2,2);
      }
      cout << num_liner_probing << endl;
      destroyTable(H2);
    
      Hashtable H3 = InitTable(200);  //200大小的散列表
      for(int i = 0; i != Num; ++i)
      {
        H3 = Insert(array_rand[i],H3,3);
      }
      cout << num_double_probing << endl;
      destroyTable(H3);
    
    
       return 0;
    }
    

      通过改变随机数组的大小,可以多次观察结果,发现每次都是双散列产生的冲突次数最少,但是也少不了多少。

         夜深了,,,

         好像是陷入死循环,希望后面的代码有个break。

         

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  • 原文地址:https://www.cnblogs.com/1242118789lr/p/6816359.html
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