Hash Function

Question

In data structure Hash, hash function is used to convert a string(or any other type) into an integer smaller than hash size and bigger or equal to zero. The objective of designing a hash function is to "hash" the key as unreasonable as possible. A good hash function can avoid collision as less as possible. A widely used hash function algorithm is using a magic number 33, consider any string as a 33 based big integer like follow:

hashcode("abcd") = (ascii(a) * 333 + ascii(b) * 332 + ascii(c) *33 + ascii(d)) % HASH_SIZE

                              = (97* 333 + 98 * 332 + 99 * 33 +100) % HASH_SIZE

                              = 3595978 % HASH_SIZE

here HASH_SIZE is the capacity of the hash table (you can assume a hash table is like an array with index 0 ~ HASH_SIZE-1).

Given a string as a key and the size of hash table, return the hash value of this key.

Clarification

For this problem, you are not necessary to design your own hash algorithm or consider any collision issue, you just need to implement the algorithm as described.

Example

For key="abcd" and size=100, return 78

Analysis

直白的实现方法就是按照定义对sum进行HASH_SIZE取模运算，但是实际上这样会产生溢出。于是应用模运算的法则，每一次累加sum时，就可以取模HASH_SIZE，这样就可以很显著地减小最终的sum值。

需要注意到是定义sum为long类型，最终返回时转化为int。

参考： http://baike.baidu.com/view/2385246.htm https://en.wikipedia.org/wiki/Modular_arithmetic

Solutions

class Solution {
    /**
     * @param key: A String you should hash
     * @param HASH_SIZE: An integer
     * @return an integer
     */
    public int hashCode(char[] key,int HASH_SIZE) {
        int N = key.length;
        long sum = 0;
        for (int i = 0; i < N; i++) {
            sum = (sum * 33 + (int) (key[i])) % HASH_SIZE;
        }

        return (int) (sum);
    }
};