使用多项式滚动散列函数字符串散列

什么是字符串哈希？

字符串散列是将字符串转换为整数的方法，称为该字符串的散列。
理想的散列是冲突可能性最小的散列（即具有相同散列的 2 个不同字符串）。

多项式滚动哈希函数

在这种散列技术中，字符串的散列计算如下：

$hash = (s[0]*P^{0} + s[1]*P^{1} + ....s[m]*P^{m}) mod M$

其中 P 和 M 是一些正数。 s[0], s[1], s[2] … s[n-1] 是分配给英文字母表中每个字符的值 (a->1, b->2, … z->26)。

P 和 M 的适当值

• P：P的值可以是大致等于使用的不同字符数的任何质数。
  例如：如果输入字符串只包含英文字母的小写字母，那么P=31就是P的合适值。
  如果输入字符串包含大写和小写字母，则 P = 53 是合适的选项。
• M：两个随机字符串碰撞的概率与m成反比，因此m应该是一个很大的素数。
  M = 10 ^9 + 9是一个不错的选择。

下面是使用多项式散列函数的字符串散列的实现：

C++

// C++ implementation of the
// Polynomial Rolling Hash Function
 
#include 
using namespace std;
 
// Function to calculate
// the hash of a string
long long polynomialRollingHash(
    string const& str)
{
    // P and M
    int p = 31;
    int m = 1e9 + 9;
    long long power_of_p = 1;
    long long hash_val = 0;
 
    // Loop to calculate the hash value
    // by iterating over the elements of string
    for (int i = 0; i < str.length(); i++) {
        hash_val
            = (hash_val
               + (str[i] - 'a' + 1) * power_of_p)
              % m;
        power_of_p
            = (power_of_p * p) % m;
    }
    return hash_val;
}
 
/// Driver Code
int main()
{
    // Given strings
    string str1 = "geeksforgeeks";
    string str2 = "geeks";
 
    cout << "Hash of '" << str1 << "' = "
         << polynomialRollingHash(str1);
    cout << endl;
}

Java

// Java implementation of the
// Polynomial Rolling Hash Function
class GFG{
 
// Function to calculate
// the hash of a String
static long polynomialRollingHash(String str)
{
     
    // P and M
    int p = 31;
    int m = (int)(1e9 + 9);
    long power_of_p = 1;
    long hash_val = 0;
 
    // Loop to calculate the hash value
    // by iterating over the elements of String
    for(int i = 0; i < str.length(); i++)
    {
        hash_val = (hash_val + (str.charAt(i) -
                    'a' + 1) * power_of_p) % m;
        power_of_p = (power_of_p * p) % m;
    }
    return hash_val;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Strings
    String str1 = "geeksforgeeks";
    String str2 = "geeks";
 
    System.out.print("Hash of '" + str1 + "' = " +
                     polynomialRollingHash(str1));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 implementation of the
# Polynomial Rolling Hash Function
 
# Function to calculate
# the hash of a string
def polynomialRollingHash(str):
     
    # P and M
    p = 31
    m = 1e9 + 9
    power_of_p = 1
    hash_val = 0
 
    # Loop to calculate the hash value
    # by iterating over the elements of string
    for i in range(len(str)):
        hash_val = ((hash_val + (ord(str[i]) -
                                 ord('a') + 1) *
                              power_of_p) % m)
 
        power_of_p = (power_of_p * p) % m
 
    return int(hash_val)
 
# Driver Code
if __name__ == '__main__':
 
    # Given string
    str1 = "geeksforgeeks"
 
    print("Hash of '{}' = {}".format(
          str1, polynomialRollingHash(str1)))
 
# This code is contributed by Shivam Singh

C#

// C# implementation of the
// Polynomial Rolling Hash Function
using System;
class GFG{
 
// Function to calculate
// the hash of a String
static long polynomialRollingHash(String str)
{   
    // P and M
    int p = 31;
    int m = (int)(1e9 + 9);
    long power_of_p = 1;
    long hash_val = 0;
 
    // Loop to calculate the hash value
    // by iterating over the elements of String
    for(int i = 0; i < str.Length; i++)
    {
        hash_val = (hash_val + (str[i] -
                    'a' + 1) * power_of_p) % m;
        power_of_p = (power_of_p * p) % m;
    }
    return hash_val;
}
 
// Driver Code
public static void Main(String[] args)
{   
    // Given Strings
    String str1 = "geeksforgeeks";
    String str2 = "geeks";
 
    Console.Write("Hash of '" + str1 + "' = " +
                   polynomialRollingHash(str1));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the
// Polynomial Rolling Hash Function
  
// Function to calculate
// the hash of a String
function polynomialRollingHash(str)
{
      
    // P and M
    let p = 31;
    let m = (1e9 + 9);
    let power_of_p = 1;
    let hash_val = 0;
  
    // Loop to calculate the hash value
    // by iterating over the elements of String
    for(let i = 0; i < str.length; i++)
    {
        hash_val = (hash_val + (str[i].charCodeAt() -
                    'a'.charCodeAt() + 1) * power_of_p) % m;
        power_of_p = (power_of_p * p) % m;
    }
    return hash_val;
}
  
 
// Driver code
 
    // Given Strings
    let str1 = "geeksforgeeks";
    let str2 = "geeks";
  
    document.write("Hash of '" + str1 + "' = " +
                     polynomialRollingHash(str1));
 
// This code is contributed by souravghosh0416.

输出：

Hash of 'geeksforgeeks' = 111226362

时间复杂度： O(N)
辅助空间：O(1)