[LintCode] Hash Function

Dogee 發(fā)布于2019-08-14 15:26 / 2591人閱讀

摘要：又用到了取余公式，推導(dǎo)出。用循環(huán)對(duì)數(shù)組各位進(jìn)行轉(zhuǎn)換和相乘和累加。因?yàn)榈诙€(gè)取余公式證明乘積取余與乘數(shù)相加后再取余等價(jià)于乘積取余，所以在每個(gè)循環(huán)內(nèi)都進(jìn)行一次取余，以免乘積太大溢出。

Problem

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

Note

又用到了取余公式： (a * k) % b = [(a % b) * (k % b)] % b，
推導(dǎo)出：(a * b) % b = (a % b + b) % b。
用for循環(huán)對(duì)char數(shù)組各位進(jìn)行hash轉(zhuǎn)換：和33相乘和累加。因?yàn)榈诙€(gè)取余公式證明乘積取余與乘數(shù)相加后再取余等價(jià)于乘積取余，所以在每個(gè)循環(huán)內(nèi)都進(jìn)行一次取余，以免乘積太大溢出。

Solution

class Solution {
    public int hashCode(char[] key,int HASH_SIZE) {
        if (key == null) return 0;
        if (HASH_SIZE == 0) return 0;
        long res = 0;
        for (int i = 0; i < key.length; i++) {
            res = res * 33 + key[i];
            res %= HASH_SIZE;
        }
        //res = (res % HASH_SIZE + HASH_SIZE) % HASH_SIZE;
        return (int)res;
    }
};