哈希表的简单操作

哈希表中，关键值通过哈希函数映射到数组上，查找时通过关键值直接访问数组。哈希表的关键问题在于哈希函数的构造和解决冲突的方法。

下面采用最简单的线性探测展示一下哈希表的基本操作：

  1 //Hashtable
  2 class HashTable {
  3 private:
  4     string *elem;
  5     int size;
  6 public:
  7     HashTable() {
  8         size = 2000;
  9         elem = new string[size];
 10         for (int i = 0; i < size; i++) {
 11             // "#" means that this slot is empty
 12             elem[i] = "#";
 13         }
 14     }
 15     ~HashTable() {
 16         delete[] elem;
 17     }
 18
 19     // get hash code of a string
 20     int hash(string &value) {
 21         int code = 0;
 22         for (size_t i = 0; i < value.length(); i++) {
 23             code = (code * 256 + value[i] + 128) % size;
 24         }
 25         return code;
 26     }
 27
 28     //search for an element
 29     //here pos and times are reference,
 30     //so in the end pos will be set to the pos of the target and times and how many slots you have tried
 31     bool search(string &value, int &pos, int &times) {
 32         pos = hash(value);
 33         times = 0;
 34         while (elem[pos] != "#" && elem[pos] != value) {
 35             times++;
 36             if (times < size) {
 37                 //test for next slot
 38                 pos = (pos + 1) % size;
 39             } else {
 40                 return false;
 41             }
 42         }
 43         if (elem[pos] == value) {
 44             return true;
 45         } else {
 46             //the slot is vacant
 47             return false;
 48         }
 49     }
 50
 51     int insert(string &value) {
 52         int pos, times;
 53         if (search(value, pos, times)) {
 54             //the elem already exists
 55             return 2;
 56         } else if (times < size / 2) {
 57             elem[pos] = value;
 58             return 1;
 59         } else {
 60             //the hashtable is too packed, need to expand and make it sparse
 61             recreate();
 62             insert(value);
 63             return 0;
 64         }
 65     }
 66
 67     void recreate(){
 68         string *temp_elem;
 69         temp_elem=new string[size];
 70         for(int i=0;i<size;i++){
 71             temp_elem[i]=elem[i];
 72         }
 73         int copy_size=size;
 74         size=size*2;
 75         delete[] elem;
 76         elem=new string[size];
 77         for(int i=0;i<size;i++){
 78             elem[i]="#";
 79         }
 80         for(int i=0;i<copy_size;i++){
 81             if(temp_elem[i]!="#"){
 82                 insert(temp_elem[i]);
 83             }
 84         }
 85         delete[] temp_elem;
 86     }
 87 };
 88
 89 int main() {
 90     HashTable hashtable;
 91     string buffer;
 92     int n;
 93     cin >> n;
 94     for (int i = 1; i <= n; i++) {
 95         cin >> buffer;
 96         int ans = hashtable.insert(buffer);
 97         if (ans == 0) {
 98             cout << "recreate while insert!" << endl;
 99         } else if (ans == 1) {
100             cout << "insert success!" << endl;
101         } else if (ans == 2) {
102             cout << "It already exists!" << endl;
103         }
104     }
105     int temp_pos, temp_times;
106     cin >> buffer;
107     if (hashtable.search(buffer, temp_pos, temp_times)) {
108         cout << "search success!" << endl;
109     } else {
110         cout << "search failed!" << endl;
111     }
112     return 0;
113 }