Hash Function

lintcode 128 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

基本实现题，大多数人看到题目的直觉是按照定义来递推，但其实这里面大有玄机，因为在字符串较长时使用long 型来计算33的幂会溢出！所以这道题的关键在于如何处理大整数溢出。对于整数求模，(a + b) % m = (a % m + b % m) % m这个基本公式务必牢记。根据这个公式我们可以大大降低时间复杂度和规避溢出。

Integer.hashCode(key)现成的hash function

public 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
        long sum = 0;
        for (int i = 0 ; i < key.length; i++) {
            sum = (sum * 33 + (int)key[i]) % HASH_SIZE;
        }
        return (int)sum % HASH_SIZE;
    }
}