Question

我正在使用线性探测制作哈希表，并且我必须在加载因子即（哈希表中输入的元素数量）/（哈希表的大小）变得大于0.5时调整数组的大小，我必须调整大小我正在通过在包含与hashtable相关的函数的类中初始化一个指针来调整大小。我将指针放在一个大小为100的struct（struct只包含一个字符串）的数组中。每次加载因子变为大于0.5，我通过创建一个前一个大小加倍的新数组来调整数组的大小，并将指针指向新数组。我还有一个int，它存储数组的当前大小，并使用调整大小函数的每个实例更新随着插入函数的每次调用，插入的元素数量都会增加。我这样做是否正确？下面是我的代码

#include <cstring>
#include <vector>
#include <math.h>
#include <iomanip>

using namespace std;

int power(int a,int b)
{
   for (int i=0;i<b;i++)
   {
       a*=a;
   }
   return a;
};

struct Bucket
{
    string word;
};   

const int size=100;

class LProbing
{
  private:
          int a; //a constant which is used in hashing
          int cursize;  //current size of hash table
          Bucket *Table;  //pointer to array of struct
          int loadfactor; //ratio of number of elements entered over size of hashtable
          int n;  //number of elements entered
          Bucket table[size];   //array of structs
  public:
          LProbing(int A);  //constant is decided by user 
          void resize();
          void insert(string word);
          void Lookup(string word);
};

LProbing::LProbing(int A)
{  
   cursize=size;                   
   a=A;
   Table=table;
   loadfactor=0;  //initially loadfactor is 0 as number of elements entered are 0
   n=0;
}

void LProbing::resize() 
{
    cout<<"resize"<<endl;
    loadfactor=n/cursize;   //ensuring if resize needs to be done
    if (loadfactor<=0.5)
    {
       return;
    }                                      
    const int s=2*cursize;  
    Bucket PTable[s];
    for (int i=0;i<cursize;i++)
    {
        if (Table[i].word.empty())
        continue;

        //rehashing the word onto the new array
        string w=Table[i].word;    
        int key=0;
        for (int j=0;j<w.size();j++)
        {
           unsigned char b=(unsigned char)w[j];
           key+=(int)power(a,i)*b;
        }
        key=key%(2*cursize);
        PTable[key].word=w;  //entering the word in the new array
    }
    Table=PTable;  //putting pointer equal to new array
    cursize=2*cursize;  //doubling the current size of array
}

void LProbing::insert(string word)
{
   cout<<"1"<<endl; 
   n++;  //incrementing the number of elements entered with every call to insert

    //if loadfactor is greater than 0.5, resize array
   loadfactor=n/cursize;
   if (loadfactor>0.5)
   {               
       resize();
    }                
   //hashing the word
   int k=0;
   for (int i=0;i<word.size();i++)
   {
       unsigned char b=(unsigned char)word[i];
       int c=(int)((power(a,i))*b);
       k+=c;
       cout<<c<<endl;
   }

   int key=0;
   key=k%cursize;
   cout<<key<<endl;
   //if the respective key index is empty enter the word in that slot
   if (Table[key].word.empty()==1)
   {
       cout<<"initial empty slot"<<endl;
       Table[key].word=word;
   }
   else  //otherwise enter in the next slot
   {
       //searching array for empty slot
       while (Table[key].word.empty()==0)
       {
        k++;
        key=k%cursize;
       }
       //when empty slot found,entering the word in that bucket
       Table[key].word=word;
       cout<<"word entered"<<endl;
   }
}             

#include "Linear Probing.cpp"
#include <fstream>

using namespace std;

int main() 
{
   LProbing H(35);
   ifstream fin;
   fin.open("dict.txt");
   vector<string> D;

   string d;
   while (getline(fin,d))
   {
       if (!d.empty())
       {
           D.push_back(d);
       }
   }
   fin.close();
   for (int i=0;i<D.size();i++)
   {
       H.insert(D[i]);
   }
   system("PAUSE");
   return 0;
}

Answer 1

您可能会发现它在某处有用： http://www.cs.rmit.edu.au/online/blackboard/chapter/05/documents/contribute/chapter/05/linear-probing.html

Answer 2

你正在处理大数字和变量“key”溢出：

key += (int)power(a,i)*b

Answer 3

看起来loadfactor计算为int/int所以它将保持为0直到达到1.尝试将分区的输入转换为浮点数。

哈希表（线性探测）

3 个答案: