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.

 

Example

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

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.

 

这道题要注意的就是溢出，要注意两点：

1. 使用long类型

2. 每次计算33的n次方要再做%HASH_SIZE，每一项累加之后也要%HASH_SIZE避免溢出

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) {        // write your code here        if(key == null || key.length == 0)              return 0;                long factor = 1;        long result = 0;                for(int i = key.length - 1; i >= 0; i--){            result += (((int) key[i]) * factor) % HASH_SIZE;            factor = (factor * 33) % HASH_SIZE;                    }                result %= HASH_SIZE;                return (int) result;    }};