Hash Function
- LintCode: No.128-Hash Function
- 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) * 33^3 + ascii(b) * 33^2 + 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.
- 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.
- Analysis:
The formula in Problem is hashcode("abcd") = (ascii(a) * 33^3 + ascii(b) * 33^2 + ascii(c) *33 + ascii(d)) % HASH_SIZE We can make it equal to hashcode("abcd") = (ascii(a) * 33^3 + ascii(b) * 33^2 + ascii(c) *33 + ascii(d) * 1) % HASH_SIZE = (ascii(a) * 33^3 + ascii(b) * 33^2 + ascii(c) *33 + ascii(d) * 33^0) % HASH_SIZE Asssume that a, b, c are integers, (a + b + c) % d = a % d + b % d + c % d. Besides, needed to concern overflow issues.
Java:
public class Solution { public int hashCode(char[] key,int HASH_SIZE) { long res = 0; for (int i = 0; i < key.length; i++){ res = 33 * res + (int)key[i]; res = res % HASH_SIZE; } return (int)res; } }
Time Complexity:
- O(n), n is key length