128.Hash Function

1.Description(Easy)--Hash

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.f

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

2.Code

基本实现题，大多数人看到题目的直觉是按照定义来递推不就得了嘛，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。

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。

(a + b) % p = (a % p + b % p) % p

(a * b) % p = (a % p * b % p) % p

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