LintCode 128. Hash Function 原创Java参考解答

LintCode 128. Hash Function 原创Java参考解答

问题描述

http://www.lintcode.com/en/problem/hash-function/

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

解题思路

题目是根据给定key数组和哈希表容量求哈希值。

  • 关键点:读题仔细,把题意转换为哈希函数公式,再把公式转换为哈希值。(前一位基础上*33+当前位置字符串ascii值) % 哈希表容量

参考代码

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 hashCode = 0; 
        //abcd 100 
        for (int i = 0; i < key.length; i++) { 
            hashCode = (hashCode * 33 + (int)(key[i])) % HASH_SIZE; 
        } 
        return (int)hashCode; 
    } 
}

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