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[tool]. Hash Function.java
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E
tags: Hash Table, Lint
time: O(1) get
space: O(n) store map
In general, there is no universal recipe to stick to when it comes to implementing hashCode().
https://www.baeldung.com/java-hashcode
#### Hash Function
- 解释Hash怎么做.
- Hash function例子:
- hashcode("abcd") = (ascii(a) * 33^3 + ascii(b) * 33^2 + ascii(c) *33^1 + ascii(d)*33^0) % HASH_SIZE
- 用到的参数比如: magic number 33, HASH_SIZE.
- Hash的意义是:给一个string key, 转换成数字,从而把size变得更小。
- 真实的implementation还要处理collision, 可能需要design hash function 等等。
##### 每一步都%HASH_SIZE的原因
- hashRst = hashRst * 33 + (int)(key[i]);
- hashRst = hashRst % HASH_SIZE;
- 原因是,hashRst会变得太大,所以不能算完和 再 %...
```
/*
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) * 33^3 + ascii(b) * 33^2 + ascii(c) *33^1 + ascii(d)*33^0) % HASH_SIZE
= (97* 33^3 + 98 * 33^2 + 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.
Example
For key="abcd" and size=100, return 78
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.
Tags Expand
Hash Table
*/
/*
Thinking process:
Use given hash function.
However, need to consider integer overflow.
A simple way: save it as a long during calculation. Then return a (int).
*/
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) {
if (key.length == 0) {
return 0;
}
long hashRst = 0;
for (int i = 0; i < key.length ; i++) {
hashRst = hashRst * 33 + (int)(key[i]);
hashRst = hashRst % HASH_SIZE;
}
return (int)hashRst;
}
};
```