[LintCode] 128 Hash Function 解题报告

Description
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


思路
题目要求照着上面的公式把结果求出来。但是我么直接做会越界。
用同余的原理。
每次搞完求一下模，这样不会越界。
注意用long来存，因为乘33可能越界。
最后变回int。


Code

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 + key[i]) % HASH_SIZE;
        }
        return (int) sum;
    }
}