LintCode HashFunction 哈希函数

在数据结构中，哈希函数是用来将一个字符串（或任何其他类型）转化为小于哈希表大小且大于等于零的整数。一个好的哈希函数可以尽可能少地产生冲突。一种广泛使用的哈希函数算法是使用数值33，假设任何字符串都是基于33的一个大整数，比如：

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

其中HASH_SIZE表示哈希表的大小(可以假设一个哈希表就是一个索引0 ~ HASH_SIZE-1的数组)。

给出一个字符串作为key和一个哈希表的大小，返回这个字符串的哈希值。

样例
对于key=”abcd” 并且 size=100， 返回 78

注意
对于这个问题你不需要自己设计hash算法，你只需要实现上述描述的hash算法即可。

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.

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) {
        long sum = (int)key[0];//sum * 33可能会超出int范围
        for(int i = 1; i < key.length; i++) {
            sum = sum * 33 % HASH_SIZE + (int)key[i];//即使求余，否则超出long范围
        }
        return (int)(sum % HASH_SIZE);
    }
}

不能写成此种形式否则超时

class Solution {
    /**
     * @param key: A String you should hash
     * @param HASH_SIZE: An integer
     * @return an integer
     */
    public static int hashCode(char[] key,int HASH_SIZE) {
        long sum = 0;
        for(int i = 0; i < key.length; i++) {
            int index = key.length - i - 1;
            long power = 1;
            while(index-- > 0){
                power = power * 33 % HASH_SIZE;
            }
            int ascii = (int)key[i];
            sum += ((ascii * power) % HASH_SIZE);
        }
        return (int)(sum % HASH_SIZE);
    }
}